A cell supplies a current of 1.2 A through two resistors each of 2 Ω connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate (i) the internal resistance and (ii) e.m.f. of the cell.

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}{Rp} = \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,

$1.2 + 1.2r = 1.6 + 0.4r \\[0.5em] 1.2r – 0.4r = 1.6 – 1.2 \\[0.5em] 0.8r = 0.4 \\[0.5em] r = \dfrac{0.4}{0.8} \\[0.5em] r = 0.5 Ω \\[0.5em]$

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