A circuit connected to an ac source of emf $$e = e_0 \sin(100t)$$ with t in seconds, gives a phase difference of $$\frac{\pi}{4}$$ between the emf $$e$$ and current $$i$$. Which of the following circuits will exhibit this?

We are told that the applied alternating emf is $$e = e_0 \sin(100t)$$, where the time $$t$$ is in seconds.

Comparing $$e = e_0 \sin(\omega t)$$ with the standard form $$e = e_0 \sin(\omega t)$$, we recognise that the angular frequency is

$$\omega = 100\ \text{rad s}^{-1}\,.$$

The circuit to be chosen must give a phase difference of

$$\phi = \frac{\pi}{4} = 45^\circ$$

between the applied emf $$e$$ and the current $$i$$. We therefore examine how the phase angle arises in series $$RC$$ and $$RL$$ circuits.

For a series RC circuit the impedance is

$$Z = R - jX_C = R - j\frac{1}{\omega C},$$

so the phase of the impedance (with respect to the real axis) is

$$\theta = -\tan^{-1}\!\!\left(\frac{X_C}{R}\right) = -\tan^{-1}\!\!\left(\frac{1}{\omega R C}\right).$$

Because the current leads the voltage by the magnitude of this angle, the current-voltage phase difference is

$$\phi_{RC} = \tan^{-1}\!\!\left(\frac{1}{\omega R C}\right).$$

For a series RL circuit the impedance is

$$Z = R + jX_L = R + j\omega L,$$

so the phase of the impedance is

$$\theta = \tan^{-1}\!\!\left(\frac{X_L}{R}\right)=\tan^{-1}\!\!\left(\frac{\omega L}{R}\right).$$

This time the current lags the voltage by the same magnitude, giving

$$\phi_{RL} = \tan^{-1}\!\!\left(\frac{\omega L}{R}\right).$$

In both cases we require the magnitude of the phase angle to be

$$\phi = \frac{\pi}{4}\,,$$

which implies

$$\tan\phi = 1.$$

We now check each option.

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

Using $$\tan\phi = \dfrac{1}{\omega R C}$$, we have

$$\tan\phi_A = \frac{1}{(100)(10^3)(10^{-6})} = \frac{1}{0.1} = 10.$$

Since $$\tan\phi_A = 10 \neq 1$$, the phase angle is not $$\pi/4$$. This option is rejected.

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

Using $$\tan\phi = \dfrac{\omega L}{R}$$, we have

$$\tan\phi_B = \frac{(100)(10^{-3})}{10^3} = \frac{0.1}{10^3} = 1\times10^{-4}.$$

Since $$\tan\phi_B \neq 1$$, the required phase difference is not obtained. This option is rejected.

Option C: RL circuit, $$R = 1\ \text{k}\Omega = 10^3\ \Omega$$, $$L = 10\ \text{mH} = 10^{-2}\ \text{H}$$.

Again using $$\tan\phi = \dfrac{\omega L}{R}$$,

$$\tan\phi_C = \frac{(100)(10^{-2})}{10^3} = \frac{1}{10^3} = 0.001.$$

This is still far from unity, so the phase angle is not $$\pi/4$$. This option is also rejected.

Option D: RC circuit, $$R = 1\ \text{k}\Omega = 10^3\ \Omega$$, $$C = 10\ \mu\text{F} = 10^{-5}\ \text{F}$$.

Using $$\tan\phi = \dfrac{1}{\omega R C}$$, we get

$$\tan\phi_D = \frac{1}{(100)(10^3)(10^{-5})} = \frac{1}{1} = 1.$$

Because $$\tan\phi_D = 1$$, we have

$$\phi_D = \tan^{-1}(1) = 45^\circ = \frac{\pi}{4}.$$

This is exactly the required phase difference. Therefore option D satisfies the given condition.

Hence, the correct answer is Option D.