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Two batteries of emf 3 V and 6 V with internal resistances $2 \Omega$ and $4 \Omega$ are connected in a circuit with resistance of $10 \Omega$ as shown in figure. The current and potential difference between the points $P$ and $Q$ are
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$\frac{3}{16} \mathrm{~A}$ and $\frac{15}{8} \mathrm{~V}$
Applying Kirchhoff's voltage law in the given loop,
$\begin{aligned}-4 I+6-3-2 I-10 I & =0 \\ -16 I & =-3 \\ I & =\frac{3}{16} \mathrm{~A}\end{aligned}$
$\because$ Potential difference across $P Q=\frac{3}{16} \times 10=\frac{15}{8} \mathrm{~V}$
$\begin{aligned}-4 I+6-3-2 I-10 I & =0 \\ -16 I & =-3 \\ I & =\frac{3}{16} \mathrm{~A}\end{aligned}$
$\because$ Potential difference across $P Q=\frac{3}{16} \times 10=\frac{15}{8} \mathrm{~V}$
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