In a series LCR circuit with value of R = 15 `Omega , L = 5 H , C = 100 muF ` an AC power supply with variable frequency is connected . Calculate the value of angular frequency of the source at which the circuit is in resonance and current at the same frequency . The `V_("rms") value of voltage in the circuit is 220 V.

Answer

To solve the problem step by step, let's break it down into two parts: finding the angular frequency at resonance and calculating the current at that frequency.

### Step 1: Calculate the Angular Frequency at Resonance

In a series LCR circuit, resonance occurs when the inductive reactance (X_L) is equal to the capacitive reactance (X_C). The formulas for these reactances are:

- Inductive Reactance: \( X_L = \omega L \)
- Capacitive Reactance: \( X_C = \frac{1}{\omega C} \)

At resonance, we have:
\[ X_L = X_C \]

This leads to:
\[ \omega L = \frac{1}{\omega C} \]

Rearranging gives:
\[ \omega^2 = \frac{1}{LC} \]

Thus, the angular frequency \( \omega \) at resonance can be calculated as:
\[ \omega = \frac{1}{\sqrt{LC}} \]

Now, substituting the given values:
- \( L = 5 \, \text{H} \)
- \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)

Calculating \( \omega \):
\[ \omega = \frac{1}{\sqrt{5 \times 100 \times 10^{-6}}} \]
\[ = \frac{1}{\sqrt{5 \times 0.0001}} \]
\[ = \frac{1}{\sqrt{0.0005}} \]
\[ = \frac{1}{0.02236} \]
\[ \approx 44.72 \, \text{rad/s} \]

### Step 2: Calculate the Current at Resonance

At resonance, the impedance \( Z \) of the circuit is equal to the resistance \( R \):
\[ Z = R = 15 \, \Omega \]

The current \( I_{rms} \) can be calculated using Ohm's law:
\[ I_{rms} = \frac{V_{rms}}{Z} \]

Given:
- \( V_{rms} = 220 \, \text{V} \)

Substituting the values:
\[ I_{rms} = \frac{220}{15} \]
\[ \approx 14.67 \, \text{A} \]

### Final Answers
1. The angular frequency at which the circuit is in resonance is approximately \( 44.72 \, \text{rad/s} \).
2. The current at this frequency is approximately \( 14.67 \, \text{A} \).