In a parallel AC circuit with R_1 = R_2 = R a | vedclass

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(D) The circuit consists of two parallel branches connected to an $AC$ source $V = V_0 \sin(\omega t)$.Branch $1$ has $R$ and $X_L$ in series. The pot

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$\frac{\sqrt{2} V_0 X R}{R^2 + X^2}$

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(D) The circuit consists of two parallel branches connected to an $AC$ source $V = V_0 \sin(\omega t)$.Branch $1$ has $R$ and $X_L$ in series. The potential at point $A$ (relative to the bottom wire) is the voltage across the inductor: $V_A = V_0 \frac{X_L}{\sqrt{R^2 + X_L^2}} \sin(\omega t + \phi)$,where $	an \phi = R/X_L$.Branch $2$ has $R$ and $X_C$ in series. The potential at point $B$ (relative to the bottom wire) is the voltage across the capacitor: $V_B = V_0 \frac{X_C}{\sqrt{R^2 + X_C^2}} \sin(\omega t - \phi)$,where $	an \phi = R/X_C$.Given $R_1 = R_2 = R$ and $X_L = X_C = X$,let $Z = \sqrt{R^2 + X^2}$. Then $V_A = \frac{V_0 X}{Z} \sin(\omega t + \phi)$ and $V_B = \frac{V_0 X}{Z} \sin(\omega t - \phi)$.The potential difference $V_A - V_B = \frac{V_0 X}{Z} [\sin(\omega t + \phi) - \sin(\omega t - \phi)] = \frac{V_0 X}{Z} [2 \cos(\omega t) \sin(\phi)]$.Since $\sin \phi = R/Z$,we have $V_A - V_B = \frac{V_0 X}{Z} [2 \cos(\omega t) \frac{R}{Z}] = \frac{2 V_0 X R}{Z^2} \cos(\omega t) = \frac{2 V_0 X R}{R^2 + X^2} \cos(\omega t)$.The peak value is $V_{peak} = \frac{2 V_0 X R}{R^2 + X^2}$. The $RMS$ value is $V_{RMS} = \frac{V_{peak}}{\sqrt{2}} = \frac{\sqrt{2} V_0 X R}{R^2 + X^2}$.

In a parallel $AC$ circuit with $R_1 = R_2 = R$ and $X_L = X_C = X$. An $AC$ source $V = V_0 \sin(\omega t)$ is connected across the circuit as shown in the figure. The $RMS$ value of the potential difference $V_A - V_B$ will be:

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