A circuit contains a resistance of `40 Omega` and an inductance of 0.68 H, and an alternating effective e.m.f. of 500 V at a frequency of 120 Hz is applied to it. Find the value of the effective current in the circuit and power factor.

Answer

To solve the problem step by step, we need to calculate the effective current in the circuit and the power factor. Here’s how to do it:

### Step 1: Identify the given values
- Resistance \( R = 40 \, \Omega \)
- Inductance \( L = 0.68 \, H \)
- Effective e.m.f. \( E_{\text{rms}} = 500 \, V \)
- Frequency \( f = 120 \, Hz \)

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 120 \approx 753.98 \, \text{rad/s} \]

### Step 3: Calculate the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]
Substituting the values of \( \omega \) and \( L \):
\[ X_L = 753.98 \times 0.68 \approx 512.71 \, \Omega \]

### Step 4: Calculate the impedance \( Z \)
The impedance \( Z \) in an R-L circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values of \( R \) and \( X_L \):
\[ Z = \sqrt{40^2 + 512.71^2} = \sqrt{1600 + 26277.54} \approx \sqrt{27877.54} \approx 167.06 \, \Omega \]

### Step 5: Calculate the effective current \( I \)
The effective current \( I \) is given by:
\[ I = \frac{E_{\text{rms}}}{Z} \]
Substituting the values of \( E_{\text{rms}} \) and \( Z \):
\[ I = \frac{500}{167.06} \approx 2.99 \, A \]

### Step 6: Calculate the power factor \( \text{pf} \)
The power factor \( \text{pf} \) is given by:
\[ \text{pf} = \cos(\phi) = \frac{R}{Z} \]
Substituting the values of \( R \) and \( Z \):
\[ \text{pf} = \frac{40}{167.06} \approx 0.239 \]

### Summary of Results
- Effective Current \( I \approx 2.99 \, A \)
- Power Factor \( \text{pf} \approx 0.239 \)