# ICSE Physics Class X (Current Electricity) Important Questions (Solved)

Question .1. Distinguish between terminal potential difference and emf of a cell.

Solution:

 Terminal potential difference Emf It is the potential of the cell when a current is drawn from it (i.e. cell in a closed circuit). It is the potential of the cell when no current is drawn from it (i.e. cell in an open circuit).

Question .2. State the energy changes occurring in an electric cell.

Solution:

Chemical energy is converted into electrical energy.

Question .3. State Ohm’s law.

Solution:

Ohm’s Law : The current flowing in a conductor is directly proportional to the potential difference across its ends provided the physical conditions and temperature of conductor remains same.

Question .4. Calculate the equivalent resistance of the following combination of resistors.

Solution:

Resistance r3 and r4 are parallel and if r be equivalent to this then

1/r = 1/r3 + 1/r4 = (r3 + r4)/r3r4 => r = r3r4/(r3 + r4).

Now resistance r1, r2 and r are in series and if R be equivalent to this then

R = r1 + r2 + r = r1 + r2 + r3r4/(r3 + r4)

= {(r1 + r2)(r3 + r4) + r3r4}/(r3 + r4).

Question .5. Write an expression for the resistance of a conducting wire in terms of its length and area of cross section.

Solution:

If R be the resistance of a wire of length L and A be its area of cross section then

R = ρL/A

where ρ is the resistivity or specific resistance of the material of the wire.

Question .6. Define the emf E of a cell and the potential difference V across a resistor R in terms of work done in moving a unit charge. State the relation between these two works and the work done in moving a unit charge through a cell connected across the resistor. Take the internal resistance of the cell as r. Hence obtain an expression for the current I in the circuit.

Solution:

The emf : The emf of a cell is equal to the work done in moving a unit positive charge from the positive terminal of the cell, through the external circuit, to the negative terminal of the cell and then in moving in it, through the electrolyte of the cell, from its negative terminal back to its positive terminal.

The potential difference : The potential difference across a resister R is equal to the work done in moving a unit positive charge from the higher potential end of the resistor to its lowest potential end. Let I be the current flowing through the circuit and the cell with internal resistance r, then the work done in moving a unit positive charge through the cell is Ir. Thus the relation between E, V, and Ir is given by

V = E – Ir

Question .7. Mention two factors on which the internal resistance of a cell depends.

Solution:

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

(i) the nature of the electrolyte used.

(ii) the area of two electrodes and the distance of separation between them.

Question .8. Mention two factors on which the resistance of a wire depends.

Solution:

The factors on which the resistance of a wire depends are :

(i) length of wire and

(ii) area of cross section of the wire.

Numerical – 1. A cell supplies a current of 0.6 A through a 2 ohm coil and a current of 0.3 A through a 8 ohm coil. Calculate the emf and the internal resistance of the cell.

Solution:

Here, I1 = 0.6 A, R1 = 2 Ω, I2 = 0.3 A, R2 = 8 Ω, r = ?

Using I = E / (R + r)

0.6 = E / (2 + r) ————– (i)

and 0.3 = E / (8 + r) ————— (ii)

From (i) and (ii) we get E = 0.6 (2 + r) = 0.3 (8 + r)

Or, 0.3 r = 1.2 => r = 4 ohm.

Substituting the value of r in (i) we get

0.6 = E / (2 + 4) => E = 0.6 (2 + 4) = 3.6 V.

Numerical – 2. Four cells each of emf 2 V and internal resistance 0.1 ohm are connected in series. The combination is connected to an ammeter of negligible resistance, a 1.6 ohm resistor and an unknown resistor R1. The current in the circuit is 2 A. (1) Draw a labelled circuit diagram for the above arrangement and (2) calculate :

(i) the total resistance of the circuit

(ii) the total emf

(iii) the value of R1 and

(iv) the potential difference across R1.

Solution:

(2) Total internal resistance = 4 × 0.1 Ω = 0.4 Ω

(i) Total resistance = 1.6 + 0.4 + R1 = (2 + R1) Ω

(ii) Total emf = (2 + 2 + 2 + 2 + 2)V = 8 V

(iii) Using I = total emf / total resistance => 2 = 8 / (2 + R1)

Or, 4 + 2 R1 = 8 => R1 = 2 Ω

(iv) Potential difference across R1 = IR1 = 2 × 2 = 4 V.

Numerical – 3. Two resistors of 2 ohm and 3 ohm in parallel are connected to a cell of emf 1.5 V and internal resistance 0.3 ohm. Draw a labelled circuit diagram showing the above arrangement and the current drawn from the cell.

Solution:

Consult a Text Book for circuit diagram.

Let r be the equivalent resistance when 2 ohm and 3 ohm are in parallel, then

1/r = 1/2 + 1/3 = 5/6 => r = 6/5 = 1.2 ohm.

Now total resistance R of the circuit is

R = r + 0.3 = 1.2 + 0.3 = 1.5 ohm.

Hence current in the circuit is

I = V / R = 1.5 / 1.5 = 1 A.

Numerical – 4. Four cells each of emf 1.5 V and internal resistance 2.0 ohm are connected in parallel. The battery of cells is connected to an external resistance of 2.5 ohm. Calculate :

(i) the total resistance of the circuit

(ii) the current flowing in the external circuit and

(iii) the drop in potential across the terminals of the cells.

Solution:

Draw the fig. from a Text Book.

(i) As cells are in parallel, the battery has net emf of 1.5 V and its net internal resistance, r is given by

1/r = ½ + ½ + ½ + ½ = 2 => r = 0.5 ohm.

(ii) total resistance, R = 0.5 + 2.5 = 3 ohm

(iii) current flowing in the external circuit is

I = 1.5 / R = 1.5 / 3.0 = 0.5 A

(iv) The potential difference across the terminals of the cells

= the potential difference across the external resistance connected to the battery.

Hence pd = V = 0.5 A × 2.5 Ω = 1.25 V

Numerical – 5. Four resistance of 2.0 Ω each are joined end to end to form a square ABCD. Calculate the equivalent resistance of the combination between any two adjacent corners.

Solution:

In this situation, three resistors are connected in series and their combination in parallel with the fourth resistor. Therefore, the series combination gives,

2 + 2 + 2 = 6 ohm.

This 6 ohm resistor is connected in parallel to 2 ohm resistor,

1/R = 1/6 + 1/2 = 2/3 => R = 3/2 = 1.5 ohm