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The terminal voltage of the battery, whose emf is 10 V and internal resistance $1 \Omega$, when connected through an external resistance of $4 \Omega$ as shown in the figure is:
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Verified Answer
The correct answer is:
8V
Current in circuit $i=\frac{10}{4+1}=2 \mathrm{~A}$
$\begin{aligned}
\text { Terminal voltage } & =E-i R \\
& =10-2 \times 1=8 \mathrm{~V}
\end{aligned}$
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