A coil of resistance 10 Ω and inductance 100mH and a capacitor of variable capacitance are connected across a 20V,50Hz A.C. supply. At what capacitance will resonance occur?

Answer

To determine the capacitance at which resonance occurs in an RLC circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance (R) = 10 Ω
- Inductance (L) = 100 mH = 100 × 10^(-3) H = 0.1 H
- Frequency (f) = 50 Hz
- Voltage (V) = 20 V (not needed for resonance calculation)

2. **Understand the Resonance Condition:**
- The resonance condition in an RLC circuit is given by the formula:
\[ \omega = \frac{1}{\sqrt{LC}} \]
- Where \( \omega \) (angular frequency) is related to frequency (f) by:
\[ \omega = 2\pi f \]

3. **Substituting for Angular Frequency:**
- Substitute \( \omega \) into the resonance condition:
\[ 2\pi f = \frac{1}{\sqrt{LC}} \]

4. **Square Both Sides:**
- Squaring both sides gives:
\[ (2\pi f)^2 = \frac{1}{LC} \]

5. **Rearranging for Capacitance (C):**
- Rearranging the equation to solve for C:
\[ C = \frac{1}{L(2\pi f)^2} \]

6. **Substituting Known Values:**
- Substitute L = 0.1 H and f = 50 Hz into the equation:
\[ C = \frac{1}{0.1 \times (2\pi \times 50)^2} \]

7. **Calculating the Values:**
- First calculate \( 2\pi \times 50 \):
\[ 2\pi \times 50 \approx 314.16 \text{ rad/s} \]
- Now square this value:
\[ (314.16)^2 \approx 98709.6 \]
- Now plug this back into the capacitance formula:
\[ C = \frac{1}{0.1 \times 98709.6} = \frac{1}{9870.96} \approx 0.0001013 \text{ F} \]

8. **Convert to Microfarads:**
- Convert the capacitance from Farads to microfarads:
\[ C \approx 101.3 \mu F \]

### Final Answer:
The capacitance at which resonance occurs is approximately **101.3 μF**.

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