An alternating e.m.f. of 200 V and 50 cycles is connected to a circuit of resistance `3.142 Omega)` and inductance 0.01 H. The lag in time between the e.m.f. and the current is

Options

A. 1.5 ms
B. 2.5 ms
C. 3.5 ms
D. None of these

Explanation

To solve the problem step by step, we need to determine the lag in time between the alternating e.m.f. and the current in the given circuit. Here’s how we can do it:

### Step 1: Identify the given values
- e.m.f (E) = 200 V
- Frequency (f) = 50 Hz
- Resistance (R) = 3.142 Ω
- Inductance (L) = 0.01 H

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 50 = 100\pi \text{ rad/s} \]

### Step 3: Calculate the inductive reactance (X_L)
Inductive reactance (X_L) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values of ω and L:
\[ X_L = (100\pi) \times 0.01 = \pi \text{ Ω} \]

### Step 4: Calculate the phase difference (φ)
The phase difference (φ) between the e.m.f. and the current can be found using the formula:
\[ \tan \phi = \frac{X_L}{R} \]
Substituting the values of X_L and R:
\[ \tan \phi = \frac{\pi}{3.142} \]
Since π is approximately equal to 3.142, we have:
\[ \tan \phi = 1 \]
Thus,
\[ \phi = \tan^{-1}(1) = 45^\circ \]

### Step 5: Calculate the time period (T)
The time period (T) is the reciprocal of the frequency:
\[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ seconds} = 20 \text{ milliseconds} \]

### Step 6: Calculate the time lag (t)
The time lag (t) can be calculated using the formula:
\[ t = \frac{\phi}{360} \times T \]
Substituting the values of φ and T:
\[ t = \frac{45}{360} \times 0.02 = \frac{1}{8} \times 0.02 = 0.0025 \text{ seconds} = 2.5 \text{ milliseconds} \]

### Conclusion
The lag in time between the e.m.f. and the current is **2.5 milliseconds**.