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(D) In the given circuit,the capacitor branch and the inductor branch are connected in parallel across the $10 \ V$ source.For the capacitor branch,th

Vedclass · Accepted Answer

$(V_{a}-V_{b})=$ zero

Vedclass · Answer

(D) In the given circuit,the capacitor branch and the inductor branch are connected in parallel across the $10 \ V$ source.For the capacitor branch,the voltage across the capacitor $V_C$ at time $t$ is given by $V_C = 10(1 - e^{-t/RC})$. Here $R = 4 \ \Omega$ and $C = 0.5 \ F$,so $RC = 4 	imes 0.5 = 2 \ s$. Thus,$V_C = 10(1 - e^{-t/2})$.The potential at point $a$ relative to the negative terminal is the voltage across the $4 \ \Omega$ resistor. Since the capacitor and resistor are in series,the current $I_C = \frac{10}{4} e^{-t/2} = 2.5 e^{-t/2}$. The voltage at $a$ is $V_a = 10 - V_C = 10 e^{-t/2}$.For the inductor branch,the current $I_L$ at time $t$ is given by $I_L = \frac{10}{2}(1 - e^{-Rt/L}) = 5(1 - e^{-2t/4}) = 5(1 - e^{-t/2})$.The potential at point $b$ relative to the negative terminal is the voltage across the inductor,$V_b = 10 - I_L R_L = 10 - 5(1 - e^{-t/2}) 	imes 2 = 10 - 10 + 10 e^{-t/2} = 10 e^{-t/2}$.Comparing the potentials,$V_a = 10 e^{-t/2}$ and $V_b = 10 e^{-t/2}$.Therefore,$(V_a - V_b) = 10 e^{-t/2} - 10 e^{-t/2} = 0$ at all times $t$.

Regarding the given circuit,the correct statement is:

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