A circuit connected to ac source of emf `e=e_(0)(100t)` with t in second gives a phase difference of `pi/4` between , emf e and current i. which of the following circuits will exhibit this?

Options

A. `RL` Circuit with `R=1 Komega` and `L=10mH`
B. `RL` circuit With `R=1 Komega` and `L=1 mH`
C. `RC` circuit With `R=1KOmega` and `C=1muF`
D. `RC` circuit with `R=1 kOmega ` and `C=10muF`

Answer

To solve the problem, we need to determine which circuit configuration will exhibit a phase difference of \(\frac{\pi}{4}\) (or 45 degrees) between the EMF \(e\) and the current \(i\). The EMF is given by \(e = e_0 \sin(100t)\).

### Step-by-step Solution:

1. **Understand the Circuit Configuration**:
- The phase difference between the voltage (EMF) and the current in an AC circuit can be influenced by the type of components in the circuit (resistor, inductor, capacitor).
- In an RL circuit, the phase difference \(\phi\) is given by \(\tan(\phi) = \frac{X_L}{R}\), where \(X_L = \omega L\) is the inductive reactance.
- In an RC circuit, the phase difference is given by \(\tan(\phi) = \frac{X_C}{R}\), where \(X_C = \frac{1}{\omega C}\) is the capacitive reactance.

2. **Calculate the Phase Difference**:
- We need to find the conditions under which the phase difference \(\phi = \frac{\pi}{4}\).
- For an RL circuit, this means:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \implies \frac{X_L}{R} = 1 \implies X_L = R \]
- For an RC circuit, this means:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \implies \frac{X_C}{R} = 1 \implies X_C = R \]

3. **Determine Values**:
- Given \(\omega = 100\) rad/s (from \(100t\)), we can express:
- For RL circuit: \(X_L = \omega L = 100L\)
- For RC circuit: \(X_C = \frac{1}{\omega C} = \frac{1}{100C}\)

4. **Check Each Option**:
- **Option A**: \(R = 1 \, \text{k}\Omega\), \(L = 10 \, \text{mH}\)
- \(X_L = 100 \times 10 \times 10^{-3} = 1 \, \Omega\) (not equal to \(R\))
- **Option B**: \(R = 1 \, \text{k}\Omega\), \(L = 1 \, \text{mH}\)
- \(X_L = 100 \times 1 \times 10^{-3} = 0.1 \, \Omega\) (not equal to \(R\))
- **Option C**: \(R = 1 \, \text{k}\Omega\), \(C = 1 \, \mu\text{F}\)
- \(X_C = \frac{1}{100 \times 1 \times 10^{-6}} = 10 \, \Omega\) (not equal to \(R\))
- **Option D**: \(R = 1 \, \text{k}\Omega\), \(C = 10 \, \mu\text{F}\)
- \(X_C = \frac{1}{100 \times 10 \times 10^{-6}} = 100 \, \Omega\) (equal to \(R\))

5. **Conclusion**:
- The only option where the phase difference is \(\frac{\pi}{4}\) is **Option D**.

### Final Answer:
**Option D** is the correct answer.