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A battery of the emf $18 \mathrm{~V}$ and internal resistance of $3 \Omega$ and another battery of emf $10 \mathrm{~V}$ and internal resistance of $1 \Omega$ are connected as shown in figure. Then, the voltmeter reading is
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Verified Answer
The correct answer is:
$12 \mathrm{~V}$
Net emf of the combination
$$
=18-10=8
$$
Net resistance of the circuit $=3+1=4$
Current in the circuit $i=\frac{8}{4}=2 \mathrm{~A}$
$$
\begin{aligned}
\therefore \text { Potential difference }(V) & =E-i r=18-6 \\
& =12 \mathrm{~V}
\end{aligned}
$$
$$
=18-10=8
$$
Net resistance of the circuit $=3+1=4$
Current in the circuit $i=\frac{8}{4}=2 \mathrm{~A}$
$$
\begin{aligned}
\therefore \text { Potential difference }(V) & =E-i r=18-6 \\
& =12 \mathrm{~V}
\end{aligned}
$$
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