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A $5 \mathrm{~V}$ battery with internal resistance $2 \Omega$ and a $2 \mathrm{~V}$ battery with internal resistance $1 \Omega$ are connected to a $10 \Omega$ resistor as shown in the figure. The current in the $10 \Omega$ resistor is
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Verified Answer
The correct answer is:
$0.03 A P_2$ to $P_1$
$0.03 A P_2$ to $P_1$
$$
\begin{aligned}
& V_{P_2}-V_{P_1}=\frac{\frac{5}{2}+\frac{0}{10}-\frac{2}{1}}{\frac{1}{2}+\frac{1}{10}+\frac{1}{1}} \\
& I=\frac{V_{P_2}-V_{P_1}}{10}=0.03 \text { from } P_2 \rightarrow P_1
\end{aligned}
$$
\begin{aligned}
& V_{P_2}-V_{P_1}=\frac{\frac{5}{2}+\frac{0}{10}-\frac{2}{1}}{\frac{1}{2}+\frac{1}{10}+\frac{1}{1}} \\
& I=\frac{V_{P_2}-V_{P_1}}{10}=0.03 \text { from } P_2 \rightarrow P_1
\end{aligned}
$$
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