The potential difference between A and B in t | vedclass

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(B) To find the potential difference between $A$ and $B$,we apply Kirchhoff's Voltage Law $(KVL)$ along the path from $A$ to $B$.Starting from point $

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$48$

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(B) To find the potential difference between $A$ and $B$,we apply Kirchhoff's Voltage Law $(KVL)$ along the path from $A$ to $B$.Starting from point $A$,the current $I = 2 \ A$ flows through the circuit.The potential at $A$ is $V_A$.Moving through the $6 \ \Omega$ resistor in the direction of current,the potential drop is $I 	imes R = 2 	imes 6 = 12 \ V$.Moving through the $12 \ V$ battery from positive to negative terminal,the potential drop is $12 \ V$.Moving through the $9 \ \Omega$ resistor,the potential drop is $2 	imes 9 = 18 \ V$.Moving through the $4 \ V$ battery from negative to positive terminal,the potential gain is $4 \ V$.Moving through the $5 \ \Omega$ resistor,the potential drop is $2 	imes 5 = 10 \ V$.Equating this to the potential at $B$ $(V_B)$:$V_A - (2 	imes 6) - 12 - (2 	imes 9) + 4 - (2 	imes 5) = V_B$$V_A - 12 - 12 - 18 + 4 - 10 = V_B$$V_A - 48 = V_B$$V_A - V_B = 48 \ V$

The potential difference between $A$ and $B$ in the following figure is (in $V$)

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