A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in figure.

Find —

(a) the reading of the ammeter,

(b) the potential difference across the terminals of the cell, and

(c) the potential difference across the 4.5 Ω resistor.

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}{Rp} = \dfrac{9 + 4.5}{4.5 \times 9} \\[0.5em] \dfrac{1}{Rp} = \dfrac{13.5}{40.5} \\[0.5em] Rp = \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] Rs = 5 Ω \\[0.5em]

Hence, the effective resistance of the circuit = 5 Ω

Using Ohm's law,

V = IR

$2 = I \times 5 \\[0.5em] I = \dfrac{2}{5} \\[0.5em] \Rightarrow I = 0.4 A$

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,

$V = 2 - (0.4 \times 1.2) \\[0.5em] V = 2 - 0.48 \\[0.5em] V = 1.52 V \\[0.5em]$

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] I1 = \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