In the circuit shown in the figure,switch S_1 | vedclass

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(B) Since switch $S_2$ is open,the capacitor is not in the circuit.The circuit consists of two parallel branches connected across the $24 \, V$ batter

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Since switch $S_2$ is open,the capacitor is not in the circuit.The circuit consists of two parallel branches connected across the $24 \, V$ battery.Branch $1$ (top): Contains $1 \, \Omega$ and $5 \, \Omega$ resistors in series. Total resistance $= 1 + 5 = 6 \, \Omega$. Current $I_1 = \frac{24}{6} = 4 \, A$.Branch $2$ (bottom): Contains two $3 \, \Omega$ resistors in series. Total resistance $= 3 + 3 = 6 \, \Omega$. Current $I_2 = \frac{24}{6} = 4 \, A$.Let the potential at the left junction be $V_O = 24 \, V$ and the right junction be $0 \, V$.For the top branch,the potential at point $b$ is $V_b = V_O - I_1 	imes 1 = 24 - (4 	imes 1) = 20 \, V$.For the bottom branch,the potential at point $a$ is $V_a = V_O - I_2 	imes 3 = 24 - (4 	imes 3) = 12 \, V$.Therefore,$V_a - V_b = 12 - 20 = -8 \, V$. The magnitude is $8 \, V$.

In the circuit shown in the figure,switch $S_1$ is initially closed and $S_2$ is open. Find $V_a - V_b$ in $V$.

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