In an ac circuit , `L = (0.4)/(pi) H` and `R = 30 Omega`. If the circuit has an alternating emf of 200 V, 50 cps, the impedance and the current in the circuit will be :

Options

A. `11.4 Omega`,17.5 A
B. `30.7 Omega`, 6.5 A
C. `40.4 Omega`, 5 A
D. `50 Omega`, 4 A

Explanation

To solve the problem step by step, we need to calculate the impedance (Z) and the current (I) in the AC circuit given the values of inductance (L), resistance (R), and the frequency of the alternating current.

### Step 1: Calculate the Inductive Reactance (X_L)

The inductive reactance (X_L) is given by the formula:
\[ X_L = \omega L \]
where \(\omega\) (angular frequency) is calculated as:
\[ \omega = 2\pi f \]
Here, \(f\) is the frequency in cycles per second (cps).

Given:
- \(L = \frac{0.4}{\pi} \, H\)
- \(f = 50 \, cps\)

First, calculate \(\omega\):
\[ \omega = 2\pi \times 50 = 100\pi \, rad/s \]

Now, substitute \(\omega\) and \(L\) into the formula for \(X_L\):
\[ X_L = (100\pi) \left(\frac{0.4}{\pi}\right) \]

The \(\pi\) cancels out:
\[ X_L = 100 \times 0.4 = 40 \, \Omega \]

### Step 2: Calculate the Impedance (Z)

The impedance (Z) in an AC circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]

Given:
- \(R = 30 \, \Omega\)
- \(X_L = 40 \, \Omega\)

Now, substitute the values into the formula:
\[ Z = \sqrt{30^2 + 40^2} \]
\[ Z = \sqrt{900 + 1600} \]
\[ Z = \sqrt{2500} \]
\[ Z = 50 \, \Omega \]

### Step 3: Calculate the RMS Current (I)

The RMS current (I) in the circuit can be calculated using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]

Given:
- \(V_{rms} = 200 \, V\)
- \(Z = 50 \, \Omega\)

Now, substitute the values:
\[ I_{rms} = \frac{200}{50} \]
\[ I_{rms} = 4 \, A \]

### Final Results

Thus, the impedance \(Z\) is \(50 \, \Omega\) and the RMS current \(I_{rms}\) is \(4 \, A\).

### Summary of Results
- Impedance (Z) = 50 Ω
- Current (I) = 4 A