For the AC circuit shown below,the phase diff | vedclass

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Question

(A) In the given circuit,the current leads the voltage,which indicates that it is an $RC$ circuit. Therefore,$P$ is a resistor and $Q$ is a capacitor.

Vedclass · Accepted Answer

$1 	ext{ k}\Omega, 10 \mu	ext{F}$

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(A) In the given circuit,the current leads the voltage,which indicates that it is an $RC$ circuit. Therefore,$P$ is a resistor and $Q$ is a capacitor.For an $RC$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + X_C^2}$.The phase difference $\phi$ is given by $	an \phi = \frac{X_C}{R}$.Given $\phi = \frac{\pi}{4}$,we have $	an(\frac{\pi}{4}) = 1$,which implies $X_C = R$.Given $Z = 1414 \Omega \approx 1000\sqrt{2} \Omega$.Substituting $X_C = R$ into the impedance formula: $Z = \sqrt{R^2 + R^2} = R\sqrt{2}$.Thus,$R\sqrt{2} = 1000\sqrt{2} \implies R = 1000 \Omega = 1 	ext{ k}\Omega$.Since $X_C = R = 1000 \Omega$ and $X_C = \frac{1}{\omega C}$,assuming $\omega = 100 	ext{ rad/s}$ (standard for such problems):$C = \frac{1}{\omega X_C} = \frac{1}{100 	imes 1000} = 10^{-5} 	ext{ F} = 10 \mu	ext{F}$.Therefore,$P = 1 	ext{ k}\Omega$ and $Q = 10 \mu	ext{F}$.

For the $AC$ circuit shown below,the phase difference between emf and current is $\frac{\pi}{4}$ radian as shown in the graph. If the impedance of the circuit is $1414 \Omega$,then the values of $P$ and $Q$ are

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