The series combination of two batteries both of the same emf 10 V , but different internal resistance of $20 Ω$ and $5 Ω$ is connected to the parallel combination of two resistors $25 Ω$ and $R Ω$ .The voltage difference across the battery of internal resistance $20 Ω$ is zero . The value of $R (in Ω)$ is,

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Detailed Solution

$F o r 10 V, 20 Ω b a t t e r y : ∴ V_{1} = ε_{1} - {ir}_{1} ⇒ 0 = 10 - i \times 20 i = 0.5 A$

$F o r 10 V, 5 Ω b a t t e r y : ∴ V_{2} = ε_{2} - {i r}_{2} = 10 - 0.5 \times 5 = 7.5 V$

$T h e p o t e n t i a l d i f f e r e n c e a c r o s s a l l t h r e e b r a n c h e s i s e q u a l .$

$V = V_{1} + V_{2} = 0 + 7.5 = 7.5 V$

$N o w, i = i_{1} + i_{2} \Rightarrow 0.5 = \frac{7.5}{25} + \frac{7.5}{R}$

$\Rightarrow 0.5 = 0.3 + \frac{7.5}{R} \Rightarrow 0.2 = \frac{7.5}{R}$

$\Rightarrow R = \frac{7.5}{0.2} = 37.5 Ω$

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