The diagram in figure shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm connected to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, and (ii) the key K is closed.

Given,

e.m.f. ε = 2 volt

resistance r = 1 ohm

external resistance R = 4 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 = 2 – (0 × 1)

V = 2 volt

Hence, voltmeter reading = 2 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,

$I = \dfrac{2}{4 + 1} \\[0.5em] I = \dfrac{2}{5} \\[0.5em] \Rightarrow I = 0.4 \text{ ampere}$

Hence, ammeter reading = 0.4 ampere

From relation,

Voltage (V) = ε – Ir

Substituting the values in the formula we get,

V = 2 - (0.4 x 1) = 1.6 V

Hence, voltmeter reading = 1.6 volt