For the network shown below, the value of V_{ | vedclass

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(C) The circuit consists of a $15 \, V$ battery in series with a $1 \, \Omega$ resistor and a $2 \, \Omega$ resistor.Total resistance $R_{eq} = 1 \, \

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$10$

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(C) The circuit consists of a $15 \, V$ battery in series with a $1 \, \Omega$ resistor and a $2 \, \Omega$ resistor.Total resistance $R_{eq} = 1 \, \Omega + 2 \, \Omega = 3 \, \Omega$.The current $i$ flowing in the circuit is given by Ohm's law: $i = \frac{V}{R_{eq}} = \frac{15 \, V}{3 \, \Omega} = 5 \, A$.To find $V_{B} - V_{A}$, we traverse the path from $A$ to $B$ through the upper branch.Starting from $A$, we go through the battery (potential increases by $15 \, V$) and then through the $1 \, \Omega$ resistor (potential decreases by $i 	imes R = 5 \, A 	imes 1 \, \Omega = 5 \, V$).Thus, $V_{B} = V_{A} + 15 \, V - 5 \, V = V_{A} + 10 \, V$.Therefore, $V_{B} - V_{A} = 10 \, V$.

For the network shown below, the value of $V_{B} - V_{A}$ is . . . . . . $V$.

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