Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of \(10 \Omega\). The internal resistances of the two batteries are \(1 \Omega\) and \(2 \Omega\) respectively. The voltage across the load lies between :

JEE Main · 2018 · 08 Apr 2018 · Offline · Current Electricity

Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of

10Ω

. The internal resistances of the two batteries are

1Ω

and

2Ω

respectively. The voltage across the load lies between :

Answer: (C)

Replace the two parallel cells by one equivalent source. The equivalent emf is the conductance-weighted average

ε_{e q} = \frac{\frac{12}{1} + \frac{13}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{18.5}{1.5} = 12.33

V, and the equivalent internal resistance is

r_{e q} = (\frac{1}{1} + \frac{1}{2})^{- 1} = \frac{2}{3} Ω

. The load then takes the share set by the voltage divider,

V = ε_{e q} \frac{R}{R + r _{e q}} = 12.33 \times \frac{10}{10 + \frac{2}{3}} \approx 11.56

V, which lies between

11.5

V and

11.6

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option