A 5 V battery with internal resistance 2,Omeg | vedclass

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(B) Let $V_{P1} = 0\,V$. Then the potential at $P_2$ is $V$. Using nodal analysis at node $P_2$:$\frac{V - 5}{2} + \frac{V}{10} + \frac{V - 2}{1} = 0$

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$0.27\,A$,$P_2$ to $P_1$

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(B) Let $V_{P1} = 0\,V$. Then the potential at $P_2$ is $V$. Using nodal analysis at node $P_2$:$\frac{V - 5}{2} + \frac{V}{10} + \frac{V - 2}{1} = 0$Multiply by $10$:$5(V - 5) + V + 10(V - 2) = 0$$5V - 25 + V + 10V - 20 = 0$$16V = 45$$V = \frac{45}{16} = 2.8125\,V$The current through the $10\,\Omega$ resistor is $I = \frac{V}{R} = \frac{2.8125}{10} = 0.28125\,A \approx 0.28\,A$.Given the options,the closest value is $0.27\,A$. Since $V_{P2} > V_{P1}$,the current flows from $P_2$ to $P_1$.

$A$ $5\ V$ battery with internal resistance $2\,\Omega$ and a $2\,V$ battery with internal resistance $1\,\Omega$ are connected to a $10\,\Omega$ resistor as shown in the figure. Find the current through the $10\,\Omega$ resistor.

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