Question 194 of 380

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

A$11.7V$ and $11.8V$B$11.6V$ and $11.7V$C$11.5V$ and $11.6V$D$11.4V$ and $11.5V$

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Answer: (C) $11.5V$ and $11.6V$

As shown in figure two cells are connected in parallel with the load resistance of $R=10\Omega$. Let current ${i}_{1},{i}_{2}$ drawn from both cells as shown, from $K.C.L.$ $i={i}_{1}+{i}_{2}...\left(1\right)$
Apply $K.V.L.$ for loop $ACDF$,
${E}_{1}-iR-{i}_{1}{r}_{1}=0 \Rightarrow 12-10i-{i}_{1}=0...\left(2\right)$
From equations $\left(1\right)$ and $\left(2\right)$
$12-11{i}_{1}-10{i}_{2}=0...\left(3\right)$
Now, for loop $BCDE$,
Apply $K.V.L.$
${E}_{2}-10i-{i}_{2}{r}_{2}=0 \Rightarrow 13-10i-2{i}_{2}=0 \Rightarrow \left(∵i={i}_{1}+{i}_{2}\right) ∴13-10{i}_{1}-12{i}_{2}=0...\left(4\right)$
From equation $\left(3\right)$ and $\left(4\right)$,
${i}_{1}=\dfrac{7}{16}A,{i}_{2}=\dfrac{23}{32}A$
So, net current in external load $R=10\Omega$ is
$I={i}_{1}+{i}_{2}=\dfrac{7}{16}+\dfrac{23}{32}=\dfrac{37}{32}A$
Then potential difference for a given load,
$V=IR=\dfrac{37}{32}\times 32=11\cdot 56V$
So, the potential difference lies between, $11\cdot 5V$ and $11\cdot 6V$.