Current Electricity

According to Ohm's law, the current flowing in a conductor is directly proportional to the potential difference applied across it's ends provided that the physical conditions and the temperature of the conductor remain constant.

If a current I flows in a conductor when the potential difference across it's ends is V and R is the resistance, then according to Ohm's law V = IR

Below circuit diagram shows the setup for verifying Ohm's law:

(a) V-I graph for a conductor obeying Ohm's law is shown below:

(b) Slope of V-I graph for a conductor represents resistance.

I-V graph for a linear resistor

Slope of I-V graph — The slope of I-V graph is $\dfrac{ΔI}{ΔV}$ which is the reciprocal of resistance of the conductor i.e.,

Slope = $\dfrac{ΔI}{ΔV}$ = $\dfrac{1}{\text{resistance of the conductor}}$

The conductors which obey the Ohm's law are called the ohmic resistors or linear resistances. Examples are all metallic conductors (such as Silver, Aluminium, Copper, Iron etc.)

Current–Voltage relationship of an ohmic resistor is shown in the below graph:

Resistance is determined in the form of slope from the graph.

The conductors which do not obey the Ohm's law are known as the non-ohmic resistors (or non-linear resistances). Examples — LED, solar cell, junction diode, etc.

Current-Voltage relationship of non-ohmic resistors is shown in the graph below:

V-I graph of a conductor at two different temperatures T₁ and T₂ is shown below:

Figure shows two straight lines A and B on the V-I graph for a conductor at two different temperatures T₁ and T₂ (T₁ > T₂) respectively.

The straight line A is more steeper than the line B because the resistance of conductor is more at high temperature T₁ than at low temperature T₂

(a) In a thick conductor, electrons get a larger area of cross section to flow as compared to a thin conductor, therefore a thick wire offers less resistance. The resistance of a conductor is inversely proportional to it's area of cross section (a), normal to the direction of flow of current i.e.,

R ∝ $\dfrac{1}{a}$

Hence,

R ∝ $\dfrac{1}{πr^2}$

where r is the radius of the wire.

Hence, resistance of a wire is inversely proportional to the square of the radius of the wire.

(b) As we have seen above that the resistance of a conductor is inversely proportional to it's area of cross section (a), normal to the direction of flow of current. Hence, the thin wire, will offer more resistance as it's area of cross section is less.

-1.6 x 10^-19 C

Reason — The charge on an electron is -1.6 x 10^-19 C.

6.25 x 10¹⁸

Reason — The number of electrons in 1 C charge is 6.25 x 10¹⁸.

ammeter, series

Reason — By connecting the ammeter in series, the entire current flowing through the circuit also flows through the ammeter, allowing it to measure the current accurately.

I = $\dfrac{ne}{t}$

Reason — If n electrons pass through a cross-section of a conductor in time t, then current in conductor is I = $\dfrac{ne}{t}$

V = W/Q

Reason — If W joule of work is done in bringing a test charge Q coulomb from infinity to the point P, then electric potential V at point P is V = W/Q .

1 joule

Reason — The potential at a point is defined as 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.

voltmeter, parallel

Reason — Connecting the voltmeter in parallel ensures that it measures the potential difference accurately because it doesn't affect the current flow in the circuit.

resistance

Reason — The obstruction offered to the flow of current by a wire is called the resistance.

All of the above

Reason — According to Ohm's Law : V = IR

Hence, I ∝ V and V/I = constant

Hence, all are correct.

volt/ampere

Reason — 1 ohm = volt/ampere

ohm^-1

Reason — The unit of conductance is ohm^-1.

solar cell

Reason — Nichrome, Copper sulphate solution with copper electrodes and dil sulphuric acid, Silver, all the three follow Ohm's Law i.e., V = IR, hence they are ohmic conductors whereas a solar cell is a non-ohmic conductor.

all of the above

Reason — The three factors on which the resistance of wire depends are —

1. Dependence on length of the wire — the resistance of a wire is directly proportional to the length of the wire.
   R ∝ l
2. Dependence on the thickness of the wire — the resistance of a wire is inversely proportional to it's area of cross section (a) normal to the direction of flow of current.
   R ∝ $\bold{\dfrac{1}{a}}$
3. Dependence on the temperature of the wire — the resistance of conductor increases with an increase in it's temperature.

Same

Reason — Specific resistance of an object is an intrinsic property which is independent of it's dimensions but does depend upon nature of material and temperature. Since both rod and thin wire are of same material i.e., aluminium so both will have same specific resistance.

9R

Reason — When the wire is stretched to thrice it's length, it's area of cross section becomes 1/3 and it's length becomes 3l.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

length = l,

Resistance = R Ω,

new length = 3l,

new area = $\dfrac{a}{3}$

From relation

R = ρ $\dfrac{l}{a}$ = ρ $\dfrac{l}{πr^2}$

Initial resistance R₁ = R = ρ $\dfrac{l}{a}$    [Equation 1]

New resistance R_n = ρ $\dfrac{3l}{\dfrac{a}{3}}$ = ρ $\dfrac{9l}{a}$    [Equation 2]

On dividing eqn (ii) by (i), we get,

$\dfrac{R_n}{R} = \dfrac{ ρ \dfrac{9l}{a}}{ρ \dfrac{l}{a}} \\[1em] \Rightarrow \dfrac{R_n}{R} = 9 \\[1em] \Rightarrow R_n = 9R$

Hence, the new resistance = 9R.

nichrome wire

Reason — The conductors which obey the Ohm's law are called the ohmic resistors or linear resistances. A Nichrome wire obeys Ohm's law hence it is an ohmic resistor.

carbon

Reason — Resistance decreases with increase in temperature for carbon.

Constantan

Reason — Standard resistors are made of constantan as it has high specific resistance and the effect of change in temperature on its resistance is negligible.

Both (1) and (3)

Reason — Nichrome wire is commonly used in heating appliances due to its following properties:

1. Specific resistance is high — Nichrome wire has a relatively high specific resistance, which means it resists the flow of electric current efficiently, resulting in heating when current passes through it.
2. Resistance increases with an increase in temperature — This property is desirable for heating elements because it helps maintain a consistent level of heating even as the temperature rises.

both (1) and (3)

Reason — A superconductor exhibits both infinite conductance and zero resistance, allowing electric current to flow through it without any loss of energy due to resistance.

Given,

Number of electrons flowing (n) = 6.25 × 10¹⁶

Time taken (t) = 2 s

e = 1.6 × 10^-19 C

Current in conductor is given by,

I = $\dfrac{\text{ne}}{\text{t}}$

Substituting the values in the formula above we get,

$I = \dfrac{6.25 \times 10^{16} \times 1.6 \times 10^{-19}}{2} \\[0.5em] I = \dfrac{6.25 \times 1.6 \times 10^{-3}}{2} \\[0.5em] I = 6.25 \times 0.8 \times 10^{-3} \\[0.5em] I = 5 \times 10^{-3} A$

Hence, the current flowing through the conductor = 5 mA from B to A

Given,

Current (I) = 3.2 mA = 3.2 × 10^-3 A

Charge of one electron = -1.6 × 10^-19 coulomb

t = 1 s

Charge flowing through the conductor in one second

Charge (Q) = current (I) x time (t)

Substituting the values in the formula above we get,

Q = 3.2 x 10^-3 x 1 = 3.2 x 10^-3

Hence,

$\text{No. of electrons} = \dfrac{\text{Total charge flowing}}{\text{Charge on one electron}}$

$= \dfrac{3.2 \times 10^{-3}}{1.6 \times 10^{-19}} \\[0.5em] = 2 \times 10^{16} \\[0.5em]$

Therefore, the number of electrons that will pass each second through the cross section of that conductor = 2 x 10¹⁶

Given,

Current (I) = 300 mA = 0.3 A

Resistance (R) = 20 Ω

Potential difference (V) = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 0.3 × 20

V = 6 V

Hence, the required potential difference = 6 V

Given,

Current (I) = 1.2 A

Potential difference (V) = 6.0 V

Resistance (R) = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

6 = 1.2 x R

⇒ R = $\dfrac{6}{1.2}$

⇒ R = 5 Ω

Hence, resistance of filament of bulb = 5 Ω

Given,

Potential difference (V) = 12 V

Current (I) = 2 A

Resistance = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

12 = 2 x R

⇒ R = $\dfrac{12}{2}$

⇒ R = 6 Ω

Hence, the resistance of the filament of the bulb = 6 Ω

The resistance will be less when the bulb is not glowing.

Given,

Resistance (R) = 5 Ohm

Potential difference (V) = 3 V

Current (I) = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

3 = I x 5

⇒ I = $\dfrac{3}{5}$

⇒ I = 0.6 A

Hence, the current flowing through the wire = 0.6 A

V-I graph for the readings is shown below:

(a) We can observe from the graph that when current is 0.5 A then potential difference is 1.25 V.

(b) We can observe from the graph that when potential difference is 0.75 V, then current is 0.3 A.

(c) Slope of the graph gives the resistance, and as the graph is linear

So, if potential difference (V) = 1.0 V

current (I) = 0.4 A

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

1.0 = 0.4 x R

⇒ R = $\dfrac{1.0}{0.4}$

⇒ R = $\dfrac{10}{4}$

⇒ R = 2.5 Ω

Hence, the resistance in circuit = 2.5 Ω

(i) R = ρ $\dfrac{l}{a}$ = ρ $\dfrac{l}{πr^2}$

Resistance for wire of radius 1 mm,

R₁ = ρ $\dfrac{l}{π1^2}$ = ρ $\dfrac{l}{π}$

Resistance for wire of radius 2 mm,

R₂ = ρ $\dfrac{l}{π2^2}$ = ρ $\dfrac{l}{π4}$

Hence, ratio between the two,

R₁ : R₂

$= \dfrac{ρ\dfrac{l}{π}}{ρ\dfrac{l}{π4}} \\[1em] = \dfrac{4}{1} \\[0.5em] = 4 : 1$

Hence, the ratio of resistance between the two wires = 4 : 1

(ii) The specific resistance of the two wires will be same because the specific resistance is a characteristic property of the material, hence it does not change with radius.

Therefore, ratio of specific resistance between the two wires = 1 : 1

When the wire is stretched to double it's length, it's area of cross section becomes half and it's length becomes double.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

length = l,

R = 1 Ω,

new length = 2l,

new area = $\dfrac{a}{2}$

From relation

R = ρ $\dfrac{l}{a}$ = ρ $\dfrac{l}{πr^2}$

Initial resistance R₁ = 1 = ρ $\dfrac{l}{a}$    [Equation 1]

New resistance R_n = ρ $\dfrac{2l}{\dfrac{a}{2}}$ = ρ $\dfrac{4l}{a}$    [Equation 2]

On dividing eqn (ii) by (i), we get,

$\dfrac{R_n}{1} = \dfrac{ ρ \dfrac{4l}{a}}{ρ \dfrac{l}{a}} \\[0.5em] \Rightarrow R_n = 4 \varOmega$

Hence, the new resistance = 4 Ω.

Given,

Resistance (R) = 3 Ω

Length (l) = 10 cm

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

R = 3 Ω

length = 10 cm,

new length = 30 cm,

new area a = $\dfrac{a}{3}$

From relation

R = ρ $\dfrac{l}{a}$ = ρ $\dfrac{l}{πr^2}$

Initial resistance 3 = ρ $\dfrac{10}{a}$    [Equation 1]

New resistance R₂ = ρ $\dfrac{30}{\dfrac{a}{3}}$ = ρ $\dfrac{90}{a}$    [Equation 2]

On dividing eqn (i) by (ii), we get,

$\dfrac{3}{R_2} = \dfrac{ρ \dfrac{10}{a}}{ ρ \dfrac{90}{a}} \\[0.5em] \dfrac{3}{R_2} = \dfrac{10}{90} \\[0.5em] \dfrac{3}{R_2} = \dfrac{1}{9} \\[0.5em] \Rightarrow R_2 = 3 \times 9 \\[0.5em] \Rightarrow R_2 = 27 \varOmega \\[0.5em]$

Hence, the new resistance = 27 Ω.

Given,

Resistance (R) = 9 Ω

When the wire is tripled on itself, it's area of cross section becomes thrice and it's length becomes $\dfrac{l}{3}$.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

length = 30 cm,

new length = $\dfrac{l}{3}$ = $\dfrac{30}{3}$ = 10 cm,

new area a_n = 3a

From relation

R = ρ $\dfrac{l}{a}$

Initial resistance 9 = ρ $\dfrac{30}{a}$    [Equation 1]

New resistance R_n = ρ $\dfrac{10}{3a}$    [Equation 2]

On dividing eqn (ii) by (i), we get,

$\dfrac{R_n}{9} = \dfrac{ρ \dfrac{10}{3a}}{ ρ \dfrac{30}{a}} \\[0.5em] \dfrac{R_n}{9} = {\dfrac{10}{90}} \\[0.5em] \dfrac{R_n}{9} = {\dfrac{1}{9}} \\[0.5em] R_n = 1 \varOmega \\[0.5em]$

Hence, the new resistance = 1 Ω.

Given,

Specific resistance ρ = 1.7 × 10^-8Ω m

Radius r = 1 mm = 10^-3m

Resistance R = 2 Ω

Length l = ?

From relation

R = ρ $\dfrac{l}{a}$ = ρ $\dfrac{l}{πr^2}$

Substituting the values in the formula above we get,

$2 = 1.7 \times 10^{-8} \dfrac{l}{π(10^{-3})^2} \\[0.5em] 2 = 1.7 \times 10^{-8} \dfrac{l}{3.14 \times (10^{-3})^2} \\[0.5em] 2 = \dfrac{1.7 \times 10^{-2} \times l}{3.14} \\[0.5em] 2 = \dfrac{1.7 \times l}{314} \\[0.5em] \Rightarrow l = \dfrac{314 \times 2}{1.7} \\[0.5em] \Rightarrow l = 369.4 \text{ m} \\[0.5em]$

Hence, the length of the copper wire = 369.4 m.

Case 1 —

Current (I) = 100 m A = 0.1 A

Potential Difference (V) = 0.2 V

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.2 = 0.1 x R

⇒ R = $\dfrac{0.2}{0.1}$

⇒ R = 2 Ω

Hence, resistance of filament of bulb = 2 Ω

Case 2 —

Current (I) = 400 m A = 0.4 A

Potentia Difference (V) = 1.0 V

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

1.0 = 0.4 x R

⇒ R = $\dfrac{1.0}{0.4}$

⇒ R = 2.5 Ω

Hence, resistance of filament of bulb = 2.5 Ω

Therefore, we observe that with increase in temperature resistance of the wire increases.

Hence, resistance of filament increases with the increase in temperature.

Current is defined as the rate of flow of charge. The S.I. unit of charge is coulomb and therefore current is measured in coulomb per second which has been given the name ampere (A).

The electric potential at a point is defined as the amount of work done per unit charge in bringing a positive test charge from infinity to that point. The S.I. unit of electric potential is volt.

The potential difference (p.d.) between two points is equal to the work done per unit charge in moving a positive test charge from one point to the other. It's S.I. unit is volt. It is a scalar quantity.

The potential difference between two points is said to be 1 volt if the work done in moving 1 coulomb charge from one point to other is 1 joule i.e.,

1 volt = $\dfrac{\text{1 joule}}{\text{1 coulomb}}$ = 1 J C^-1

(a) Current is a scalar quantity.

By stating the direction of current in a conductor we mean that the direction of motion of electrons is opposite to it.

(b) Potential is a scalar quantity.

When the potential is positive at a point in the vicinity of a positive charge since work has to be done on the positive test charge against the repulsive force due to the positive charge in bringing it from infinity, while it is negative at a point in the vicinity of a negative charge since the attractive force on test charge due to the negative charge does work by itself.

The obstruction offered to the flow of current by the conductor (or wire) is called it's resistance.

The S.I. unit of resistance is Ohm.

(a) The particles which are responsible for the flow of current in a metallic wire are free electrons.

(b) In a metallic wire, the moving charges are the free electrons which constitute the current.

If n electrons pass through the cross section of a conductor in time t, then total charge passed through the conductor is given as

Q (charge) = n × e

and the current in conductor is

$\text{I} = \dfrac{\text{Q}}{\text{t}} = \dfrac{\text{ne}}{\text{t}}$

(c) When the ends of a metal wire are connected to a cell, the electrons inside it experience a force in direction from the end at negative potential to positive, due to which they began to move as shown in the diagram below.

Due to force, their speed increases but during the movement they collide with the fixed positive ions and loose some of their kinetic energy due to which their speed decreases. This lost energy heats up the wire.

After the collision, they are again accelerated towards the positive potential due to the existing potential difference so their speeds again increases and then again in collision with the positive ions, their speed decreases. This process continues. As a result, the electrons do not move in bulk with a continuously increasing speed, but their is a drift of electrons towards the positive terminal. Thus, a metal wire offers some resistance to the flow of electrons through it.

(a) The law which relates the potential difference and current in a conductor is known as Ohm's law.

Ohm's law states that the current flowing through the conductor is directly proportional to the potential difference across it's ends provided that the physical conditions and the temperature of the conductor remain constant.

(b) The necessary condition for a conductor to obey Ohm's law is that the physical conditions and the temperature should remain constant.

Graph (a) is non-ohmic resistor and

Graph (b) is ohmic resistor.

The I-V graph for (b) is a straight line or linear while for (a), the graph is a curve.

Resistance of a wire is directly proportional to the length of the wire.

R ∝ l

In a long conductor, the number of collisions of free electrons with the positive ions will be more as compared to a shorter one. Therefore, a longer conductor offers more resistance. Hence, the resistance of a conductor is directly proportional to the length of the wire.

With the increase in temperature of a conductor, the random motion of electrons increases. As a result, the number of collisions of electrons with the positive ions increase. Hence, the resistance of a conductor increases with an increase in it's temperature.

The iron wire will have more resistance.

Iron wire has more resistance as compared to copper because specific resistance of iron is more than that of copper.

The three factors on which the resistance of wire depends are —

1. Dependence on length of the wire — the resistance of a wire is directly proportional to the length of the wire.
   R ∝ l
2. Dependence on the thickness of the wire — the resistance of a wire is inversely proportional to it's area of cross section (a) normal to the direction of flow of current.
   R ∝ $\bold{\dfrac{1}{a}}$
3. Dependence on the temperature of the wire — the resistance of conductor increases with an increase in it's temperature.

Specific resistance of a material is the resistance of a wire of that material of unit length and unit area of cross section.

The S.I. unit of specific resistance is ohm × metre or (Ω m)

(a) Two factors on which the specific resistance of a wire depends are —

1. Material of the substance — It is a characteristic property of the substance. It is different for different substances.
2. Temperature of the substance — It increases with the increase in temperature for metals, but it decreases with the increase in temperature for the semiconductor.

(b) Both the wires will have the same specific resistance because the specific resistance is a characteristic property of the material and as both the wires are of copper hence both will have same specific resistance.

The specific resistance of a semi-conductor decreases with the increase in temperature.

(a) Dependency of resistance of a wire is as follows:

1. Resistance of a wire is directly proportional to length of the wire.
2. Resistance of a wire is inversely proportional to the square of radius of the wire.

(b) Dependency of specific resistance of a wire is as follows:

1. Specific resistance of a wire does not depend on length of the wire as it is a characteristic property of the material of wire.
2. Specific resistance of a wire does not depend on radius of the wire as it is a characteristic property of the material of wire.

(a) Copper or aluminium materials are used for making connection wires because their specific resistance is very low and hence they possess least possible resistance.

Due to low (or negligible) resistance of connection wires, the current in circuit remains unaffected, and the loss of energy due to heating is prevented. Hence, they are made of materials such as copper or aluminium, whose specific resistance is very small.

(b) The connection wires are made thick so that their resistance becomes low.

Standard resistors are made of magnanin, constantan etc. for which the specific resistance is high and the effect of change in temperature on their resistance is negligible.

The material used for making fuse wire is an alloy of lead and tin because it's melting point is low and it's specific resistance is more than that of copper or aluminium so that the resistance of a short and thin fuse wire is more than that of the connecting wire. It permits current upto it's safe limit to pass through it, but an excessive current melts it so that it blows off and the circuit is broken.

A superconductor is a substance of zero resistance (or infinite conductance) at a very low temperature.

Example — Mercury below 4.2 K, lead below 7.25 K and niobium below 9.2 K

Resistances of these substances decrease tremendously with the decrease in temperature and become almost zero in the low temperature range near absolute zero. Zero resistance of a superconductor means it's infinite conductivity (i.e., once a current starts flowing in a superconductor, it persists even when there is no potential across it).

The expression is —

R = ρ $\dfrac{ l}{a}$

where

R = resistance of wire
ρ = specific resistance of the material of wire
l = length of wire
a = area of cross section of wire

The order of specific resistance is as follows:

1. The specific resistance is very low for metals, because it allows most of current to pass through it.
2. The specific resistance is low for semiconductor.
3. The specific resistance is very high for insulators, as the current won't pass through it.

The substance whose specific resistance remains almost unchanged with increase in temperature is manganin.

(i) The material used for the filament of an electric bulb is a tungsten wire because it has a high melting point.

(ii) The material used for the heating element of a room heater is nichrome because the specific resistance of nichrome is high and it's resistance increases to a great extent with the increase in temperature.

A substance of zero resistance (or infinite conductance) at a very low temperature is called a superconductor.

e.m.f — e.m.f of a cell is defined as the energy spent (or the work done) per unit charge in taking a positive test charge around the complete circuit of the cell. (i.e., in the circuit outside the cell as well as in the electrolyte inside the cell).

Terminal voltage — The terminal voltage of a cell is defined as the work done per unit charge in carrying a positive test charge around the circuit connected across the terminals of the cell.

Internal resistance — The resistance offered by the electrolyte inside the cell, to the flow of current, is known as the internal resistance of the cell.

In a closed circuit, the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. Hence, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.

Hence, the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.

Below diagram shows two resistors connected in series:

If current I is drawn from the battery, the current through each resistor will also be I.

By Ohm's law,

p.d. between A and B is V₁ = V_A - V_B = IR₁

p.d. between B and C is V₂ = V_B - V_C = IR₂

Adding these we get,

V = V₁ + V₂

= V_A - V_B + V_B - V_C

= V_A - V_C

= IR₁ + IR₂

= I (R₁ + R₂)    [Equation 1]

If the equivalent resistance between the points A and C is R_S, then the potential difference between the points A and C is

V = V_A - V_C = IR_S    [Equation 2]

Therefore from equation 1 and 2,

IR_S = I (R₁ + R₂)

⇒ R_S = R₁ + R₂

Thus, in the series combination, the equivalent resistance is equal to the sum of the individual resistances.

Below diagram shows two resistors connected in parallel:

Let I₁ and I₂ be the currents through the resistances R₁ and R₂ respectively, then total current drawn from the battery is

I = I₁ + I₂    [Equation 1]

If potential difference between the two ends A and B is V, then by Ohm's law

current in R₁ is I₁ = $\dfrac{V}{R_1}$

current in R₂ is I₂ = $\dfrac{V}{R_2}$

On adding these,

I₁ + I₂ = $\dfrac{V}{R_1}$ + $\dfrac{V}{R_2}$    [Equation 2]

If the equivalent resistance of the combination between the points A and C is R_p, then total current drawn from the source is

I = $\dfrac{V}{R_p}$    [Equation 3]

Substituting the values of I and I₁ + I₂ from equation 3 and 2 in 1, we get,

$\dfrac{V}{R_p} = V\Big(\dfrac{1}{R_1} + \dfrac{1}{R_2}\Big) \\[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$

Thus, in the parallel combination, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

the material of electrodes

Reason — The emf of a cell depends on :

1. the material of electrodes and
2. the electrolyte used in the cell.

equal to, greater than

Reason — Emf of a cell is equal to the terminal voltage when cell is not in use and when the current starts to flow in the circuit, then there is a voltage drop due to internal resistance, thus we get, V = E - Ir. Hence, Emf is greater than the terminal voltage when cell is in use.

I = $\dfrac{{\varepsilon}}{{(R + r)}}$

Reason — The current drawn from the cell

I = $\dfrac{\text { e.m.f. of cell}}{\text { total resistance}}$ = $\dfrac{{\varepsilon}}{{(R + r)}}$

current is same in each resistance

Reason — Current is same in each resistance in series combination of resistances. The current has a single path to flow, hence, same current passes through each resistor.

p.d. is the same across each resistance

Reason — Potential difference is same across each resistor in parallel combination of resistances, which is equal to the potential difference across the terminals of the battery (or source).

All of the above

Reason — For parallel combination of resistances :

1. On I-V graph, the slope of line is more because resistance in parallel combination is less than series combination.
2. The potential difference across each resistance is same because, the ends of each resistor are connected to the ends of the same source of potential.
3. According to Ohm's Law V = IR. Hence, the current in a resistor is inversely proportional to the resistance.

assertion is false but reason is true..

Explanation

Assertion (A) is false. E.m.f. is greater than the terminal voltage only when the cell delivers the current to the external circuit.

Reason (R) is true. When a heavy current is drawn from the cell, a large number of charge carriers flow through the electrolyte and hence more work is done. This results in more voltage drop v, and hence less terminal voltage V. Thus, terminal voltage V of a cell depends on the amount of current I drawn from it.

Given,

e.m.f. ε = 4 volt

resistance r = 2 ohm

external resistance R = 8 ohm

(i) When the key is open then no current is flowing in the circuit and hence the ammeter reading = 0

From relation,

Voltage (V) = ε – Ir

Substituting the values in the formula we get,

V = 4 – (0 × 2)

V = 4 volt

Hence, voltmeter reading = 4 volt

(ii) When key is closed, current drawn from the cell,

$\text{I} = \dfrac{\text{e.m.f. of cell}}{\text{total resistance}} = \dfrac{\text{ε}}{\text{R} + \text{r}}$

Substituting the values in the formula above we get,

I = $\dfrac{4}{8 + 2}$

⇒ I = $\dfrac{4}{10}$

⇒ I = 0.4 ampere

Hence, ammeter reading = 0.4 ampere

From relation,

Voltage (V) = ε – Ir

Substituting the values in the formula we get,

V = 4 - (0.4 x 2) = 3.2 V

Hence, voltmeter reading = 3.2 volt

Given,

e.m.f. (ε) = 6.0 V

Current (I) = 3 A

Potential difference (V) = 5.4 V

r = ?

From relation,

V = ε – Ir

Substituting the values in the formula above we get,

5.4 = 6 - (3r)

⇒ r = 6 - 5.4

⇒ 3r = 0.6

⇒ r = $\dfrac{0.6}{3}$

⇒ r = $\dfrac{6}{30}$

⇒ r = $\dfrac{2}{10}$

⇒ r = 0.2 Ω

Hence, internal resistance of the battery = 0.2 Ω

(a) Given,

e.m.f. (ε) = 1.8 V

Internal resistance (r) = 2 Ω

I = ?

Total resistance of arrangement = 2 + 0.7 + 4.5 = 7.2 Ω

From relation,

$I = \dfrac{ε}{R}$

Substituting the values in the formula above we get,

I = $\dfrac{1.8}{7.2}$

⇒ I = 0.25 A

Hence, reading of ammeter = 0.25 A

(b) Current (I) = 0.25 A

total resistance (excluding internal resistance) = 4.5 + 0.7 = 5.2 ohm

Using ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 × 5.2

V = 1.3 V

Hence, potential difference across the terminals of the battery = 1.3 V

(a) Given,

I = 400 mA = 400 x 10^-3 = 0.4 A

V = 12 V

From Ohm's law,

V = IR

Substituting the values in the formula we get,

12 = 0.4 x R

⇒ R = $\dfrac{12}{0.4}$

⇒ R = 30 Ω

Hence, the resistance of the music system = 30 Ω

(b) Given,

I = 320 mA = 320 x 10^-3 A = 0.32 A

R = 30 Ω

From Ohm's law,

V = IR

Substituting the values in the formula we get,

V = 0.32 x 30 = 9.6V

Hence, the battery voltage when the music system stops playing = 9.6 V

Given,

e.m.f. = ε

internal resistance = r

current (I) = 1.0 A

external resistance (R) = 1.9 ohm

Case 1

From relation,

ε = I (R + r)

Substituting the value in the formula above we get,

ε = 1 (1.9 + r)

ε = 1.9 + r    [Equation 1]

In second case,

I = 0.5 A,

R = 3.9 Ω

Substituting the value in the formula above we get,

ε = 0.5 (3.9 + r)

ε = 1.95 + 0.5r    [Equation 2]

Equating 1 and 2 we get,

$1.9 + r = 1.95 + 0.5r \\[0.5em] r - 0.5 r = 1.95 - 1.9 \\[0.5em] 0.5 r = 0.05 \\[0.5em] \Rightarrow r = \dfrac{0.05}{0.5} \\[0.5em] \Rightarrow r = 0.1 Ω$

Now, substituting the value of r in equation 1, we get,

ε = 1.9 + 0.1

ε = 2 V

Hence, ε = 2 V , r = 0.1 Ω

In parallel, if the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{8} + \dfrac{1}{12} \\[0.5em] \dfrac{1}{R_P} = \dfrac{3 +2}{24} \\[0.5em] \dfrac{1}{R_P} = \dfrac{5}{24} \\[0.5em] R_P = \dfrac{24}{5} \\[0.5em] \Rightarrow R_P = 4.8 Ω$

Hence, equivalent resistance = 4.8 Ω

Give,

R₁ = R₂ = R₃ = R₄ = 5 Ω

In parallel, if the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} \\[1em] \dfrac{1}{R_P} = \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} \\[1em] \dfrac{1}{R_P} = \dfrac{4}{5} \\[1em] \Rightarrow R_P = \dfrac{5}{4} \\[1em] \Rightarrow R_P = 1.25 Ω$

Hence, equivalent resistance = 1.25 Ω

In order to get a total resistance less than 1 Ω, the three resistors of values 2 Ω, 3 Ω and 5 Ω, should be connected in parallel as shown in the diagram below:

In parallel, if the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{5} \\[0.5em] \dfrac{1}{R_P} = \dfrac{15 + 10+ 6}{30} \\[0.5em] \dfrac{1}{R_P} = \dfrac{31}{30} \\[0.5em] \Rightarrow R_P = 0.97 Ω$

Hence, the three resistors should be connected in parallel and total resistance = 0.97 Ω

When two resistors (2 Ω each) in parallel combination are connected to the third resistor (2 Ω) in series connection then the resultant resistance is 3 Ω. The below diagram shows this arrangement of resistors:

To verify,

Total resistance of two resistors in parallel,

In parallel, if the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{2} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{2}{2} \\[0.5em] \Rightarrow R_P = 1 Ω$

Hence, R_P = 1 Ω

In series connection,

Equivalent resistance = R_E = R_P + R₃
= 1 + 2 = 3 Ω

Hence, Equivalent resistance = 3 Ω

Let the resistors be R₁, R₂, R₃, R₄ each of 2.0 Ω.

In parallel, if the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{R_3} + \dfrac{1}{R_4} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{2} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{2}{2} \\[0.5em] \Rightarrow R_P = 1 Ω$

Hence, R_P = 1 Ω

In series connection,

Equivalent resistance = R_E = R₁ + R₂ + R_P
= 2 + 2 + 1 = 5 Ω

Hence, Equivalent resistance = 5 Ω

Let the four set of resistors be R₁, R₂, R₃, R₄ .

As each set consists of three resistors of 2 Ω each in series:
Resistance of each set = 2 + 2 + 2 = 6 Ω

As four similar sets of equivalent resistance 6 Ω are arranged in parallel, so the equivalent resistance (R_P) will be:

$\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_P} = \dfrac{4}{6} \\[0.5em] R_P = \dfrac{6}{4} \\[0.5em] \Rightarrow R_P = 1.5 Ω$

Hence, equivalent resistance = 1.5 Ω

In the circuit, there are three parts. In the first part, resistors of 4 Ω and 8 Ω are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = 4 + 8 = 12 Ω

In the second part, resistors of x Ω and 5 Ω are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (x + 5) Ω

In the third part, the two parts of resistance R'_s = 12 Ω and R''_s = (x + 5) Ω are connected in parallel. If the equivalent resistance between points A and B is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{R'_s} + \dfrac{1}{R''_s} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{12} + \dfrac{1}{x + 5} \\[0.5em]$

But it is given that equivalent resistance between points A and B is 4 Ω
∴ R_p = 4 Ω

Putting value of R_p in the equation above and solving for x:

$\dfrac{1}{4} = \dfrac{1}{12} + \dfrac{1}{x + 5} \\[0.5em] \dfrac{1}{4} = \dfrac{x + 5 +12}{12(x+5)} \\[0.5em] \dfrac{1}{4} = \dfrac{x + 17}{12(x+5)} \\[0.5em] \dfrac{12(x+5)}{4} = x + 17 \\[0.5em] 3(x+5) = x + 17 \\[0.5em] 3x + 15 = x + 17 \\[0.5em] 3x - x = 17 - 15 \\[0.5em] 2x = 2 \\[0.5em] \Rightarrow x = 1 Ω$

Hence, the value of x = 1 Ω

In the circuit, there are four parts. In the first part, three resistors of 1 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (1 + 1 + 1) Ω = 3 Ω

In the second part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (2 + 2 + 2) Ω = 6 Ω

In the third part, the two parts of resistance R'_s = 3 Ω and R''_s = 6 Ω and 2 Ω are connected in parallel. If the equivalent resistance is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{R'_s} + \dfrac{1}{2} + \dfrac{1}{R''_s} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{2} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2 + 3 + 1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{6}{6} \\[0.5em] \Rightarrow R_p = 1 Ω \\[0.5em]$

∴ R_p = 1 Ω

In the fourth part 1 Ω, (R_p = 1 Ω) and 1 Ω are connected in series in between points A and B.

Hence, the equivalent resistance is 1 + 1 + 1 = 3 Ω

Given, uniform wire with a resistance of 27 Ω is divided into three equal pieces. Hence, resistance of each piece = 9 Ω.

Three such pieces are joined in parallel.

In parallel, if the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{9} + \dfrac{1}{9} + \dfrac{1}{9} = \dfrac{3}{9} = \dfrac{1}{3} \\[0.5em] \Rightarrow R_P = 3 Ω$

Hence, equivalent resistance = 3 Ω

Below diagram shows the arrangement of resistors:

In the circuit, there are two parts. In the first part, two resistors of 6 Ω and 12 Ω are connected in parallel. If the equivalent resistance is R_p then

$\dfrac{1}{R_P} = \dfrac{1}{12} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1+2}{12} = \dfrac{3}{12} = \dfrac{1}{4} \\[0.5em] \Rightarrow R_P = 4 Ω$

Hence, R_P = 4 Ω

In the second part, 1 Ω and (R_P = 4 Ω) are connected in series. If the equivalent resistance is R_s then

R_s = 1 + R_P = 1 + 4 = 5 Ω

Hence, total resistance of the circuit = 5 Ω

In the circuit, there are two parts. In the first part, resistors of 12 Ω, 6 Ω and 4 Ω are connected in parallel. If the equivalent resistance of this part is R'_p then

$\dfrac{1}{{R'}_P} = \dfrac{1}{12} + \dfrac{1}{6} + \dfrac{1}{4} \\[0.5em] \dfrac{1}{{R'}_P} = \dfrac{1 + 2 + 3}{12} \\[0.5em] \dfrac{1}{{R'}_P} = \dfrac{6}{12} \\[0.5em] \dfrac{1}{{R'}_P} = \dfrac{1}{2} \\[0.5em] \Rightarrow {R'}_P = 2 Ω$

Hence, R'_P = 2 Ω

In the second part, resistors of 2 Ω, (R'_p = 2Ω) and 5 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = (2 + 2 + 5) Ω = 9 Ω

Hence, Effective resistance between A and B = 9 Ω

In the circuit, there are three parts. In the first part, two resistors of 3 Ω, 2 Ω are connected in series. If the equivalent resistance is R'_s then

R'_s = 3 + 2 = 5 Ω

In the second part, two resistors of 6 Ω, 4 Ω are connected in series. If the equivalent resistance is R''_s then

R''_s = 6 + 4 = 10 Ω

In the third part R'_s, 30 Ω and R''_s are connected in parallel. If the equivalent resistance is R_p then

$\dfrac{1}{R_P} = \dfrac{1}{R'_s} + \dfrac{1}{30} + \dfrac{1}{R''_s} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{5} + \dfrac{1}{30} + \dfrac{1}{10} \\[0.5em] \dfrac{1}{R_P} = \dfrac{6 + 1 + 3}{30} \\[0.5em] \dfrac{1}{R_P} = \dfrac{10}{30} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{3} \\[0.5em] \Rightarrow R_P = 3 Ω$

Hence, R_P = 3 Ω

In the circuit, there are two parts. In the first part, three resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (2 + 2 + 2) Ω = 6 Ω

In the second part, the resistance of first part (R'_s = 6 Ω ) and 2 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 3}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{4}{6} \\[0.5em] \Rightarrow R_p = 1.5 Ω \\[0.5em]$

∴ R_p = 1.5 Ω

(b) In the circuit, there are three parts. In the first part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (2 + 2) Ω = 4 Ω

In the second part, two resistors of 2 Ω each are connected in series. If the equivalent resistance of this part is R''_s then

R''_s = (2 + 2) Ω = 4 Ω

In the third part, the two parts of resistance R'_s = 4 Ω and R''_s = 4 Ω are connected in parallel. If the equivalent resistance between points A and C is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{4} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2}{4} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{2} \\[0.5em] \Rightarrow R_p = 2 Ω \\[0.5em]$

∴ Equivalent Resistance between A and C = 2 Ω

(a) In the circuit, there are two parts. In the first part, two resistors of 3 Ω each are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (3 + 3) Ω = 6 Ω

In the second part, resistance R'_s = 6 Ω and 3 Ω are connected in parallel. If the equivalent resistance between points P and Q is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 2}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3}{6} \\[0.5em] \Rightarrow R_p = 2 Ω \\[0.5em]$

∴ Equivalent resistance between the points P and Q = 2 Ω

(b) In the circuit, 3 Ω, R_p = 2 Ω and 3 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = (3 + 2 + 3) Ω = 8 Ω

∴ Equivalent resistance between the points X and Y = 8 Ω

(a) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in series with a battery of 6.0 V and negligible internal resistance is shown below:

Given,

Two resistors of 4 Ω and 6 Ω are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (4 + 6) Ω = 10 Ω

Potential Difference V = 6 V

Current I = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

6 = I x 10
⇒ I = 6 / 10 = 0.6 A

Hence, in series, current through the battery = 0.6 A

(b) Circuit diagram showing two resistors of 4.0 Ω and 6.0 Ω connected in parallel with a battery of 6.0 V and negligible internal resistance is shown below:

Given,

Two resistors of 4 Ω and 6 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3 + 2}{12} = \dfrac{5}{12} \\[0.5em] R_p = \dfrac{12}{5} = 2.4 Ω$

Potential Difference V = 6 V

Current I = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

6 = I x 2.4 ⇒ I = 6 / 2.4 = 2.5 A

Hence, in parallel, current through the battery = 2.5 A

(a) Given,

Two resistors of 6 Ω and 4 Ω are connected in series. If the equivalent resistance of this part is R'_s then

R'_s = (6 + 4) Ω = 10 Ω

Potential Difference V = 20 V

Current I = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

20 = I x 10
⇒ I = 20 / 10 = 2 A

Hence, in series, current through the battery = 2 A

(b) Given,

Resistance R = 6 Ω

Potential Difference V = ?

Current I = 2 A

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 2 x 6 = 12 V

Hence, potential difference = 12 V

(a) Labelled diagram of the arrangement is shown below:

(b) Two resistors of 2 Ω and 3 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{3} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3 + 2}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{5}{6} \\[0.5em] R_p = \dfrac{6}{5} \\[0.5em] \Rightarrow R_p = 1.2 Ω$

Hence, equivalent resistance = 1.2 Ω

Current I = 0.5 A

Potential Difference V = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

V = 0.5 x 1.2 = 0.6 V

Current through 2 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.6 = I x 2 ⇒ I = 0.6 / 2 = 0.3 A

Hence, current through 2 Ω resistor = 0.3 A

Current through 3 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

0.6 = I x 3 ⇒ I = 0.6 / 3 = 0.2 A

Hence, current through 3 Ω resistor = 0.2 A

(a) Given,

R = 1 Ω

V = 2 V

Current through 1 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

2 = I x 1
⇒ I = 2 / 1 = 2 A

Hence, current through Resistor A = 2 A

(b) Given,

R = 2 Ω

V = 2 V

Current through 2 Ω = ?

From Ohm's law

V = IR

Substituting the values in the formula above, we get,

2 = I x 2
⇒ I = 2 / 2 = 1 A

Hence, current through Resistor B = 1 A

(a) Given,

V = 4 V

I = 0.4 A

From Ohm's law

V = IR

4 = 0.4 x R
⇒ R = 4 / 0.4 = 10 Ω

Hence, total resistance of the circuit = 10 Ω

(b) As total resistance of the circuit is equal to 10 Ω and the two resistors R and 20 Ω are connected in parallel, hence we get,

$\dfrac{1}{10} = \dfrac{1}{R} + \dfrac{1}{20} \\[0.5em] \dfrac{1}{R} = \dfrac{1}{10} - \dfrac{1}{20} \\[0.5em] \dfrac{1}{R} = \dfrac{2 - 1}{20} \\[0.5em] \dfrac{1}{R} = \dfrac{1}{20} \\[0.5em] R = 20 Ω \\[0.5em]$

Hence, the value of R = 20 Ω

(c) current flowing in R = ?

V = 4 V

R = 20 Ω

I = ?

From Ohm's law

V = IR
⇒ 4 = I x 20
⇒ I = 4 / 20 = 1 / 5 = 0.2 A

Hence, the current flowing in R = 0.2 A

(a) Resistance of 1 m of wire = 3 Ω.

Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 Ω. As three such wires are joined in parallel and if the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{4.5} + \dfrac{1}{4.5} + \dfrac{1}{4.5} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3}{4.5} \\[0.5em] \Rightarrow R_p = \dfrac{4.5}{3} \\[0.5em] \Rightarrow R_p = 1.5 Ω$

Hence, total resistance of circuit = 1.5 Ω

(b) I = 2 A

From Ohm's Law

V = IR

Substituting the values in the formula above we get,

V = 2 x 4.5 = 9 V

Hence, potential difference = 9 V

(c) R = 3 Ω for 1 meter wire

Therefore, for 5 m

R = 3 x 5 = 15 Ω

Here the area is twice and resistance is inversely proportional to area.

Thus, resistance becomes half

R = 15 / 2 = 7.5 Ω

Hence, resistance = 7.5 Ω

(i) Given,

Two resistors each of 2 Ω connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2}{2} \\[0.5em] \Rightarrow R_p = 1 Ω$

Hence, equivalent resistance R_p = 1 Ω

Given,

I = 1.2 A

From relation,

ε = I (R + r)

ε = 1.2 (1 + r)

ε = 1.2 + 1.2r    [Equation 1]

When the resistors are connected in series, it supplies a current of 0.4 A,

If the equivalent resistance of this part is R_s then

R_s = 2 + 2 = 4 Ω

Hence, equivalent resistance R_s = 4 Ω

Given, I = 0.4 A

From relation,

ε = I (R + r)

ε = 0.4 (4 + r)

ε = 1.6 + 0.4r    [Equation 2]

Equating 1 and 2, we get,

1.2 + 1.2r = 1.6 + 0.4r

⇒ 1.2r – 0.4r = 1.6 - 1.2

⇒ 0.8r = 0.4

⇒ r = $\dfrac{0.4}{0.8}$

⇒ r = 0.5 Ω

Hence, Internal resistance r = 0.5 Ω

(ii) Substituting the value in equation 1 we get,

ε = 1.2(1 + r)
= 1.2 (1 + 0.5)
= 1.2 x 1.5
= 1.8 V

Hence, e.m.f. of the cell = 1.8 V

(a) Given,

e.m.f. = 16 V

internal resistance r = 2 Ω

current through battery = ?

If R_p is the equivalent resistance of resistors 3 Ω and 6 Ω connected in parallel, then

$\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2 + 1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{3}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1}{2} \\[0.5em] \Rightarrow R_p = 2 Ω$

From relation,

ε = I (R + r)

Substituting the value in the formula above we get,

16 = I(2 + 2)
⇒ 16 = I x 4 ⇒ I = 16 / 4 = 4 A

Hence, current through the battery = 4 A

(b) Potential difference between the terminals of the battery = ?

Using Ohm's law

V = IR

R = 2 Ω

I = 4 A

Substituting the values in the formula above we get,

V = 4 x 2 = 8 V

Hence, potential difference between the terminals of the battery = 8 V

(c) Current in 3 Ω resistor = ?

Using Ohm's law

V = IR

R = 3 Ω

V = 8 V

I = ?

Substituting the values in the formula above we get,

8 = I × 3

I = $\dfrac{8}{3}$

⇒ I = 2.66 A

Hence, current in 3 Ω resistor is 2.66 A

(d) Current in 6 Ω resistor = ?

Using Ohm's law

V = IR

R = 6 Ω

V = 8 V

I = ?

Substituting the values in the formula above we get,

8 = I × 6
⇒ I = 8 / 6 = 1.333 A

Hence, current in 6 Ω resistor is 1.34 A

(a) Given,

resistor = 4 Ω

I = 0.25 A

the p.d. across the 4 Ω resistor (V) = ?

Using Ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 x 4 = 1 V

Hence, the p.d. across the 4 Ω resistor (V) = 1 V

(b) Given,

internal resistance = 3 Ω

I = 0.25 A

the p.d. across the internal resistance of the cell = ?

Using Ohm's law

V = IR

Substituting the values in the formula above we get,

V = 0.25 x 3 = 0.75 V

Hence, the p.d. across the internal resistance (V) = 0.75 V

(c) Potential difference across R Ω or 2 Ω

V = V_net - V_{across 4 Ω} - V_{across 3 Ω}

Hence, we get,

V = 2 - 1 - 0.75 = 0.25 V

(d) The p.d. across resistor of R Ω = 0.25 V

Let the equivalent resistance of the resistors of 2 Ω and R Ω connected in parallel be R'_p

$\dfrac{1}{R'_p} = \dfrac{1}{R} + \dfrac{1}{2} \\[0.5em] \Rightarrow \dfrac{1}{R'_p} = \dfrac{2 + R}{2R} \\[0.5em] \Rightarrow R'_p = \dfrac{2R}{2 + R} \\[0.5em]$

Using Ohm's law

V = IR

0.25 = 0.25 x R'_p

Substituting the value of R'_p from above:

0.25 = 0.25 x $\Big(\dfrac{2\text{R}}{2 + \text{R}}\Big)$

⇒ $\dfrac{2\text{R}}{2 + \text{R}}$ = 1

⇒ 2R = 2 + R

⇒ 2R - R = 2

⇒ R = 2 Ω

Hence, value of R = 2 Ω

(a) In the circuit, there are two parts. In the first part, resistors of 2.0 and 4.0 Ω are connected in series. If the equivalent resistance of this part is R_s then

R_s = 2 + 4 = 6 Ω

In the second part, R_s = 6.0 and resistor of 6.0 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{1 + 1}{6} \\[0.5em] \dfrac{1}{R_p} = \dfrac{2}{6} \\[0.5em] R_p = \dfrac{6}{2} \\[0.5em] R_p = 3.0 Ω \\[0.5em]$

Hence, the effective resistance of the circuit = 3 Ω

(b) The reading of ammeter = ?

R = 3 Ω

V = 6.0 V

Using Ohm's law,

V = IR

Substituting the values in the formula above we get,

6 = I x 3
⇒ I = 6 / 3 = 2 A

Hence, the reading of ammeter = 2 A

(a) In the circuit, there are three parts. In the first part, resistors of 10 Ω and 40 Ω are connected in parallel. If the equivalent resistance of this part is R'_p then

$\dfrac{1}{R'_p} = \dfrac{1}{10} + \dfrac{1}{40} \\[0.5em] \dfrac{1}{R'_p} = \dfrac{4 + 1}{40} \\[0.5em] \dfrac{1}{R'_p} = \dfrac{5}{40} \\[0.5em] R'_p = \dfrac{40}{5} \\[0.5em] R'_p = 8 Ω \\[0.5em]$

In the second part, resistors of 30 Ω, 20 Ω and 60 Ω are connected in parallel. If the equivalent resistance of this part is R''_p then

$\dfrac{1}{R''_p} = \dfrac{1}{30} + \dfrac{1}{20} + \dfrac{1}{60} \\[0.5em] \dfrac{1}{R''_p} = \dfrac{2 + 3 + 1}{60} \\[0.5em] \dfrac{1}{R''_p} = \dfrac{6}{60} \\[0.5em] R''_p = \dfrac{60}{6} \\[0.5em] R''_p = 10 Ω \\[0.5em]$

In the third part, resistors R'_p and R''_p are connected in series. If the equivalent resistance of this part is R_s then

$R_s = 8 + 10 \\[0.5em] R_s = 18 Ω \\[0.5em]$

Hence, the total resistance of the circuit = 18 Ω

(b) Given,

e.m.f. = 1.8 V

effective resistance of the circuit = 18 Ω

I = ?

From Ohm's law

V= IR

Substituting the values in the formula above, we get,

1.8 = I x 18

⇒ I = $\dfrac{1.8}{18}$

⇒ I = 0.1 A

Hence, the reading of ammeter = 0.1 A

(a) Given,

e.m.f. = 2V

I = ?

In the circuit, there are two parts. In the first part, resistors of 4.5 Ω and 9 Ω are connected in parallel. If the equivalent resistance of this part is R_p then

$\dfrac{1}{R_p} = \dfrac{1}{4.5} + \dfrac{1}{9} \\[0.5em] \dfrac{1}{R_p} = \dfrac{9 + 4.5}{4.5 \times 9} \\[0.5em] \dfrac{1}{R_p} = \dfrac{13.5}{40.5} \\[0.5em] R_p = \dfrac{40.5}{13.5} \\[0.5em] R_p = 3 Ω \\[0.5em]$

In the second part, 1.2 Ω, 0.8 Ω and R_p = 3 Ω are connected in series. If the equivalent resistance of this part is R_s then

$R_s = 1.2 + 0.8 + 3 \\[0.5em] R_s = 5 Ω \\[0.5em]$

Hence, the effective resistance of the circuit = 5 Ω

Using Ohm's law,

V = IR

2 = I x 5

⇒ I = $\dfrac{2}{5}$

⇒ I = 0.4 A

Hence, the reading of the ammeter = 0.4 A

(b) The potential difference across the ends of the cells = ?

ε = 2 V

I = 0.4 A

r = 1.2 Ω

From relation,

Voltage (V) = ε – Ir

Substituting the values in the formula we get,

V = 2 - (0.4 x 1.2)

⇒ V = 2 - 0.48

⇒ V = 1.52 V

Hence, potential difference across the terminals of the cell = 1.52 V

(c) The potential difference across the 4.5 Ω resistor = ?

Current flowing is I = 0.4 A. Now the current I divides in two parts. Let the current in 4.5 Ω resistor be I₁ and in 9 Ω resistor be I₂.

So I = I₁ + I₂

and I₁ x 4.5 = I₂ x 9

On solving,

$I_1 = \dfrac{9}{4.5 + 9} \times I \\[0.5em] I_1 = \dfrac{9}{4.5 + 9} \times 0.4 \\[0.5em] I_1 = 0.2667 \\[0.5em]$

p.d. across the 4.5 Ω resistor

$= I_1 \times 4.5 \\[0.5em] = 0.2667 \times 4.5 \\[0.5em] = 1.2 V$

Alternate Method:

V_4.5Ω = V_cell - V_ammeter

$V_{4.5Ω} = 1.52 - IR \\[0.5em] V_{4.5Ω} = 1.52 - (0.4 \times 0.8) \\[0.5em] V_{4.5Ω} = 1.52 - 0.32 \\[0.5em] V_{4.5Ω} = 1.2 V$

Hence, p.d. across the 4.5 Ω resistor = 1.2 V

The factors on which the internal resistance of a cell depends are —

1. The surface area of the electrodes — larger the surface area of electrodes, less is the internal resistance.
2. The distance between the electrodes — more the distance between the electrodes, greater is the internal resistance.

Given,

e.m.f. = ε

internal resistance = r

external resistance = R

(a) The total resistance of circuit = R + r

(b) The current drawn from the cell

I = $\dfrac{\text { e.m.f.of cell}}{\text { total resistance}}$ = $\dfrac{\bold{\varepsilon}}{\bold{(R + r)}}$

(c) the p.d. across the cell = V = IR

Substituting from above,

V = $\bold{\dfrac{\varepsilon}{(R + r)}}$ x R

(d) voltage drop inside the cell V = Ir

Substituting from above,

V = $\bold{\dfrac{\varepsilon}{(R + r)}}$ x r

(a) When a cell is used to send current to an external circuit, it's terminal voltage V is less than it's e.m.f.

Hence, terminal voltage < e.m.f.

(b) The emf of the cell is equal to it's terminal voltage when no current is drawn.

The slope of V-I graph gives the resistance. Since the straight line A is less steeper than B, so the straight line A represents small resistance. In parallel combination, the equivalent resistance is less than in series combination so A represents the parallel combination.

(a) Equivalent resistance R of three resistors R₁, R₂ and R₃ joined in parallel is given by —

$\dfrac{1}{R}$ = $\dfrac{1}{R}_1$ + $\dfrac{1}{R_2}$ + $\dfrac{1}{R_3}$

(b) Equivalent resistance R of three resistors R₁, R₂ and R₃ joined in series is given by —

R = R₁ + R₂ + R₃

(a) When same current flows in each resistor, then the two resistors are joined in series

(b) When potential difference is same across each resistor, then the two resistors are joined in parallel

(c) When equivalent resistance is less than either of the two resistances, then the two resistors are joined in parallel

(d) When equivalent resistance is more than either of the two resistances, then the two resistors are joined in series

(a) In the circuit, two resistors of 2 Ω each are connected in parallel. In parallel, the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{2} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{2}{2} \\[0.5em] \Rightarrow R_P = 1 Ω$

Hence, R_P = 1 Ω

(b) In the circuit, three resistors of 1 Ω, 1 Ω and 2 Ω are connected in parallel. In parallel, the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{2 + 2 + 1}{2} \\[0.5em] \Rightarrow R_P = \dfrac{2}{5} Ω \\[0.5em] \Rightarrow R_P = 0.4 Ω$

Hence, R_P = 0.4 Ω

(c) In the circuit two resistors of 1 Ω each are connected in parallel. In parallel, the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{1} + \dfrac{1}{1} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1 + 1}{1} \\[0.5em] \dfrac{1}{R_P} = \dfrac{2}{1} \\[0.5em] R_P = \dfrac{1}{2} \\[0.5em] \Rightarrow R_P = 0.5 Ω$

Hence, R_P = 0.5 Ω

(d) In the circuit there are three parts. In the first part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R'_S, then

R'_S = (1 + 1) Ω = 2 Ω

In the second part two resistor of 1 Ω each are connected in series. In series, the equivalent resistance is R''_S, then

R''_S = (1 + 1) Ω = 2 Ω

In the third part,R'_S and R''_S are in parallel, the equivalent resistance is R_P, then

$\dfrac{1}{R_P} = \dfrac{1}{2} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1 + 1}{2} \\[0.5em] \dfrac{1}{R_P} = \dfrac{2}{2} \\[0.5em] \Rightarrow R_P = 1 Ω$

Hence, R_P = 1 Ω

Therefore, on observing the values we can say that (a) and (d) have same equivalent resistances i.e., 1 Ω.

(i) Energy consumed by fluorescent tube in one day = 0.24 kWh

Hence, energy consumed in 30 days = 0.24 x 30 = 7.2 kWh.

(ii) Energy consumed by television set in one day = 0.36 kWh

Hence, energy consumed in 30 days = 0.36 x 30 = 10.8 kWh.

(iii) Energy consumed by room heater in one day = 2 kWh

Hence, energy consumed in 30 days = 2 x 30 = 60 kWh.

Resistance of lamp A (rating 220 V, 50 W) is —

$R_A = \dfrac{V^2}{P_A}$