Question 222 of 380

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

A$0.27 A P_2$ to $P_1$B$0.03 A P_1$ to $P_2$C$0.03 A P_2$ to $P_1$D$0.27 \mathrm{~A} \mathrm{P} P_1$ to $P_2$

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Answer: (C) $0.03 A P_2$ to $P_1$

$$\begin{aligned} & V_{P_2}-V_{P_1}=\frac{\frac{5}{2}+\frac{0}{10}-\frac{2}{1}}{\frac{1}{2}+\frac{1}{10}+\frac{1}{1}} \\ & I=\frac{V_{P_2}-V_{P_1}}{10}=0.03 \text { from } P_2 \rightarrow P_1 \end{aligned}$$