In the given circuit,find V_a - V_b. | vedclass

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(D) Let the potential at the left junction be $V_L = 0 	ext{ V}$ and the potential at the right junction be $V_R = 10 	ext{ V}$.The circuit consists

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(D) Let the potential at the left junction be $V_L = 0 	ext{ V}$ and the potential at the right junction be $V_R = 10 	ext{ V}$.The circuit consists of two parallel branches connected across the $10 	ext{ V}$ source.Branch $1$ (left side): The resistors $2 \ \Omega$ and $3 \ \Omega$ are in series. The potential at point $a$ is determined by the voltage divider rule: $V_a = V_L + (V_R - V_L) 	imes \frac{2}{2+3} = 0 + 10 	imes \frac{2}{5} = 4 	ext{ V}$.Branch $2$ (right side): The resistors $4 \ \Omega$ and $6 \ \Omega$ are in series. The potential at point $b$ is determined by the voltage divider rule: $V_b = V_L + (V_R - V_L) 	imes \frac{4}{4+6} = 0 + 10 	imes \frac{4}{10} = 4 	ext{ V}$.Thus,$V_a - V_b = 4 	ext{ V} - 4 	ext{ V} = 0 	ext{ V}$.

In the given circuit,find $V_a - V_b$.

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