In a series L.C.R. circuit, the potential drop across `L`, `C` and `R` respectively are `40V`, `120V` and `60V` . Them the source voltage is

Options

A. 220 V
B. 160 V
C. 180 V
D. 100 V

Answer

To find the source voltage in a series LCR circuit where the potential drops across the inductor (L), capacitor (C), and resistor (R) are given as 40V, 120V, and 60V respectively, we can follow these steps:

### Step-by-Step Solution

1. **Identify the given values:**
- Voltage across the inductor (VL) = 40V
- Voltage across the capacitor (VC) = 120V
- Voltage across the resistor (VR) = 60V

2. **Understand the phase relationships:**
- In an LCR circuit, the voltage across the inductor (VL) leads the current by 90 degrees.
- The voltage across the capacitor (VC) lags the current by 90 degrees.
- The voltage across the resistor (VR) is in phase with the current.

3. **Set up the phasor diagram:**
- The voltage across the resistor (VR) is along the horizontal axis (real axis).
- The voltage across the inductor (VL) is represented as a vertical vector pointing upwards (imaginary axis).
- The voltage across the capacitor (VC) is represented as a vertical vector pointing downwards (imaginary axis).

4. **Calculate the net voltage across the inductor and capacitor:**
- The effective voltage across the inductor and capacitor can be calculated as:
\[ V_{LC} = V_L - V_C = 40V - 120V = -80V \]
- Since this is a phasor addition, we consider the magnitude:
\[ |V_{LC}| = 80V \]

5. **Use the Pythagorean theorem to find the source voltage (V):**
- The source voltage is given by the vector sum of the voltages across the resistor and the effective voltage across the inductor and capacitor:
\[ V = \sqrt{V_R^2 + V_{LC}^2} \]
- Substituting the values:
\[ V = \sqrt{(60V)^2 + (80V)^2} \]
\[ V = \sqrt{3600 + 6400} = \sqrt{10000} = 100V \]

6. **Conclusion:**
- The source voltage is 100V.

### Final Answer:
The source voltage is **100V**.

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