An inductor `2//pi` Henry, a capacitor `100//T muF` and a resistor `75Omega` are connected in series across a source of emf `V=10sin 100pit`. Here t is in second. (a) find the impedance of the circuit. (b) find the energy dissipated in the circuit in 20 minutes.

Answer

To solve the problem step by step, we will break it down into parts (a) and (b) as given in the question.

### Part (a): Finding the Impedance of the Circuit

1. **Identify the given values:**
- Inductance, \( L = \frac{2}{\pi} \) Henry
- Capacitance, \( C = \frac{100}{\pi} \, \mu F = \frac{100}{\pi} \times 10^{-6} \, F \)
- Resistance, \( R = 75 \, \Omega \)
- Voltage source, \( V = 10 \sin(100 \pi t) \)

2. **Determine the angular frequency \( \omega \):**
- From the voltage equation, we have \( \omega = 100 \pi \, \text{rad/s} \).

3. **Calculate the inductive reactance \( X_L \):**
\[ X_L = \omega L = (100 \pi) \left(\frac{2}{\pi}\right) = 200 \, \Omega \]

4. **Calculate the capacitive reactance \( X_C \):**
\[ X_C = \frac{1}{\omega C} = \frac{1}{100 \pi \left(\frac{100}{\pi} \times 10^{-6}\right)} = \frac{10^{-6}}{100} = 100 \, \Omega \]

5. **Calculate the impedance \( Z \):**
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
\[ Z = \sqrt{75^2 + (200 - 100)^2} = \sqrt{75^2 + 100^2} = \sqrt{5625 + 10000} = \sqrt{15625} = 125 \, \Omega \]

### Part (b): Finding the Energy Dissipated in the Circuit in 20 Minutes

1. **Calculate the peak current \( I_0 \):**
\[ I_0 = \frac{V_0}{Z} = \frac{10}{125} = 0.08 \, A \]

2. **Calculate the RMS current \( I_{RMS} \):**
\[ I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{0.08}{\sqrt{2}} = \frac{0.08 \times \sqrt{2}}{2} = 0.05657 \, A \]

3. **Calculate the energy dissipated \( E \):**
\[ E = I_{RMS}^2 \cdot R \cdot t \]
- Convert time from minutes to seconds: \( t = 20 \times 60 = 1200 \, s \)
\[ E = (0.05657)^2 \cdot 75 \cdot 1200 \]
\[ E = 0.0032 \cdot 75 \cdot 1200 = 288 \, J \]

### Final Answers:
- (a) The impedance of the circuit is \( Z = 125 \, \Omega \).
- (b) The energy dissipated in the circuit in 20 minutes is \( E = 288 \, J \).