Two batteries of $e.m.f.$ $4\,V$ and $8 \,V$ with internal resistances $1\, \Omega$ and $2\,\Omega$ are connected in a circuit with a resistance of $9 \,\Omega$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are
AIPMT 1988, Medium
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Applying Kirchoff's voltage law in the given loop.
$ - 2i + 8 - 4 - 1 \times i - 9i = 0$ $ \Rightarrow $ $i = \frac{1}{3}A$
Potential difference across $PQ$ $ = \frac{1}{3} \times 9 = 3\,V$
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