1. The applied emf is $e=e_0\sin(100t)$ So the angular frequency is $\omega=100\ \text{rad s}^{-1}$

2. For a series $RC$ or $RL$ circuit, the phase angle magnitude is given by:

• For $RC$: $\tan\phi=\frac{1}{\omega CR}$ (Current leads voltage)
• For $RL$: $\tan\phi=\frac{\omega L}{R}$ (Current lags voltage)

Given phase difference is $\phi=\frac{\pi}{4}$ Hence, $\tan\phi=1$

So we test each option.

3. Option A: $RC$ circuit with $R=1\,\text{k}\Omega=1000\,\Omega$, $C=1\,\mu F=10^{-6}F$

\tan\phi=\frac{1}{\omega CR}=rac{1}{100\times 10^{-6}\times 1000} $=\frac{1}{0.1}=10$ So, $\phi\ne \frac{\pi}{4}$ Therefore, A is incorrect.

4. Option B: $RL$ circuit with $R=1000\,\Omega$, $L=1\,mH=10^{-3}H$

$\tan\phi=\frac{\omega L}{R}=\frac{100\times 10^{-3}}{1000}$ $=\frac{0.1}{1000}=10^{-4}$ So, $\phi\ne \frac{\pi}{4}$ Therefore, B is incorrect.

5. Option C: $RC$ circuit with $R=1000\,\Omega$, $C=10\,\mu F=10^{-5}F$

\tan\phi=\frac{1}{\omega CR}=rac{1}{100\times 10^{-5}\times 1000} $=\frac{1}{1}=1$ Hence, $\phi=\frac{\pi}{4}$ Therefore, C is correct.

6. Option D: $RL$ circuit with $R=1000\,\Omega$, $L=10\,mH=10^{-2}H$

$\tan\phi=\frac{\omega L}{R}=\frac{100\times 10^{-2}}{1000}$ $=\frac{1}{1000}=10^{-3}$ So, $\phi\ne \frac{\pi}{4}$ Therefore, D is incorrect.

7. Therefore, the circuit that gives phase difference $\pi/4$ is: $\boxed{\text{C}}$

8. Comparison with stored correct answer:

• Stored correct answer: C
• Derived answer: C

They agree.