As shown in the figure,a 5,V battery with 2,O | vedclass

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(C) Let the current from the $5\,V$ battery be $I$ and the current through the $10\,\Omega$ resistor be $I_1$. By Kirchhoff's junction rule at node $P

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

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(C) Let the current from the $5\,V$ battery be $I$ and the current through the $10\,\Omega$ resistor be $I_1$. By Kirchhoff's junction rule at node $P_2$,the current through the $2\,V$ battery branch is $I_2 = I - I_1$.Applying Kirchhoff's voltage law $(KVL)$ to the left loop $(AD P_2 P_1 D)$: $5 - 2I - 10I_1 = 0 \implies 2I + 10I_1 = 5$.Substituting $I = I_1 + I_2$: $2(I_1 + I_2) + 10I_1 = 5 \implies 12I_1 + 2I_2 = 5$ (Equation $1$).Applying $KVL$ to the right loop $(P_2 B C P_1 P_2)$: $10I_1 - 1I_2 - 2 = 0 \implies 10I_1 - I_2 = 2$ (Equation $2$).Multiply Equation $2$ by $2$: $20I_1 - 2I_2 = 4$ (Equation $3$).Adding Equation $1$ and Equation $3$: $(12I_1 + 2I_2) + (20I_1 - 2I_2) = 5 + 4 \implies 32I_1 = 9 \implies I_1 = 9/32 = 0.28125\,A \approx 0.27\,A$.Since $I_1$ is positive,the current flows from $P_2$ to $P_1$.

As shown in the figure,a $5\,V$ battery with $2\,\Omega$ internal resistance and a $2\,V$ battery with $1\,\Omega$ internal resistance are connected to a $10\,\Omega$ resistor. The current flowing through the $10\,\Omega$ resistor is:

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