In a series R-C circuit shown in the figure,t | vedclass

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(A) For a series $R-C$ circuit,the total applied voltage $V$ is given by the phasor sum of the voltage across the resistor $V_R$ and the voltage acros

Vedclass · Accepted Answer

$6\, V, 	an^{-1}\left(\frac{4}{3}ight)$

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(A) For a series $R-C$ circuit,the total applied voltage $V$ is given by the phasor sum of the voltage across the resistor $V_R$ and the voltage across the capacitor $V_C$:$V^2 = V_R^2 + V_C^2$Given $V = 10\, V$ and $V_C = 8\, V$,we have:$10^2 = V_R^2 + 8^2$$100 = V_R^2 + 64$$V_R^2 = 100 - 64 = 36$$V_R = 6\, V$Now,the phase difference $\phi$ between the current and the applied voltage is given by:$	an \phi = \frac{V_C}{V_R}$$	an \phi = \frac{8}{6} = \frac{4}{3}$$\phi = 	an^{-1}\left(\frac{4}{3}ight)$Thus,the voltage across $R$ is $6\, V$ and the phase difference is $	an^{-1}\left(\frac{4}{3}ight)$.

In a series $R-C$ circuit shown in the figure,the applied voltage is $10\, V$ and the voltage across the capacitor is found to be $8\, V$. Then,the voltage across $R$ and the phase difference between the current and the applied voltage will respectively be

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