An inductor is connected in series with an capacitor and a resistor to an AC supply of V volts. The readings in voltmeters connected across inductor capacitor and resistor are 120 V, 50 V and 80 V respectively . Calculate the value of V and hence also find the power factor.

Answer

To solve the problem, we need to calculate the total voltage (V) supplied to the circuit and the power factor of the circuit. The circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series. The voltages across each component are given as follows:

- Voltage across the inductor (V_L) = 120 V
- Voltage across the capacitor (V_C) = 50 V
- Voltage across the resistor (V_R) = 80 V

### Step 1: Calculate the resultant voltage (V)

To find the total voltage (V) supplied to the circuit, we can use the concept of vector addition of voltages in an AC circuit. The voltages across the inductor and capacitor are out of phase by 180 degrees (they are anti-parallel), while the voltage across the resistor is in phase with the total voltage.

1. **Calculate the net voltage across the inductor and capacitor:**
\[ V_{LC} = V_L - V_C = 120V - 50V = 70V \]

2. **Now, we have a right triangle formed by the voltages:**
- One side is \( V_{LC} = 70V \) (net voltage across L and C)
- The other side is \( V_R = 80V \) (voltage across the resistor)

3. **Using Pythagoras' theorem to find the total voltage (V):**
\[ V = \sqrt{(V_{LC})^2 + (V_R)^2} \]
\[ V = \sqrt{(70)^2 + (80)^2} = \sqrt{4900 + 6400} = \sqrt{11300} \approx 106.3V \]

### Step 2: Calculate the power factor (cos φ)

The power factor (PF) is defined as the cosine of the angle (φ) between the total voltage and the current in the circuit. In a series circuit, the power factor can be calculated using the formula:
\[ \text{Power Factor} = \cos φ = \frac{V_R}{V} \]

1. **Substituting the values:**
\[ \text{Power Factor} = \cos φ = \frac{80V}{106.3V} \approx 0.75 \]

### Final Results

- The total voltage \( V \) supplied to the circuit is approximately **106.3 V**.
- The power factor of the circuit is approximately **0.75**.