For the AC circuit shown, the phase differenc | vedclass

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(A) $1$. 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 ca

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

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(A) $1$. 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.$2$. For an $RC$ circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + X_C^2}$.$3$. 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 $X_C = R$.$4$. Given $Z = 1414 \Omega \approx 1000\sqrt{2} \Omega$. Substituting $X_C = R$ into the impedance formula: $Z = \sqrt{R^2 + R^2} = R\sqrt{2}$.$5$. Thus, $R\sqrt{2} = 1000\sqrt{2} \implies R = 1000 \Omega = 1 	ext{ k}\Omega$.$6$. Since $X_C = R = 1000 \Omega$ and $X_C = \frac{1}{\omega C}$, assuming $\omega = 100 	ext{ rad/s}$ (standard for such problems): $C = \frac{1}{\omega X_C} = \frac{1}{100 	imes 1000} = 10^{-5} 	ext{ F} = 10 \mu	ext{F}$.$7$. Therefore, $P = 1 	ext{ k}\Omega$ and $Q = 10 \mu	ext{F}$.

For the $AC$ circuit shown, the phase difference between emf and current is $\frac{\pi}{4} \text{ rad}$ 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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