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

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

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(A) In a series $CR$ 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 $(V_C)$: $V = \sqrt{V_R^2 + V_C^2}$ Given $V = 10 \, V$ and $V_C = 8 \, V$. Substituting these values: $10 = \sqrt{V_R^2 + 8^2}$ $100 = V_R^2 + 64$ $V_R^2 = 36 \implies V_R = 6 \, V$. The phase difference $\phi$ between the current and the applied voltage in a $CR$ circuit is given by: $	an \phi = \frac{V_C}{V_R} = \frac{8}{6} = \frac{4}{3}$ Therefore,$\phi = 	an^{-1} \left( \frac{4}{3} ight)$. Thus,the voltage across $R$ is $6 \, V$ and the phase difference is $	an^{-1} \left( \frac{4}{3} ight)$.

In a series $CR$ 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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