A circuit is set up by connecting L=100mH, C`=5muF` and R`=100Omega` in series. An alternating emf of 150` sqrt(2)V,(500)/(pi) Hz is applied across this series combination. Which of the following is correct:

Options

A. the impedance of the circuit is `141.4 Omega`
B. the average power dissipated across resistance `225 W`
C. the average power dissipated across inductor is zero.
D. the average power dissipated across capacitor is zero

Explanation

To solve the problem step by step, we will analyze the given circuit parameters and calculate the required values.

### Step 1: Identify the given values
- Inductance (L) = 100 mH = 100 × 10^-3 H
- Capacitance (C) = 5 µF = 5 × 10^-6 F
- Resistance (R) = 100 Ω
- Alternating emf (V) = 150√2 V
- Frequency (f) = (500/π) Hz

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \left(\frac{500}{\pi}\right) = 1000 \, \text{rad/s} \]

### Step 3: Calculate the inductive reactance (X_L)
The inductive reactance (X_L) is given by:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 1000 \times (100 \times 10^{-3}) = 100 \, \Omega \]

### Step 4: Calculate the capacitive reactance (X_C)
The capacitive reactance (X_C) is given by:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{1000 \times (5 \times 10^{-6})} = \frac{1}{0.000005} = 200 \, \Omega \]

### Step 5: Calculate the impedance (Z)
The total impedance (Z) in a series RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{100^2 + (100 - 200)^2} = \sqrt{10000 + (-100)^2} = \sqrt{10000 + 10000} = \sqrt{20000} = 141.42 \, \Omega \]

### Step 6: Calculate the current (I)
The RMS current (I) can be calculated using Ohm's law:
\[ I = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I = \frac{150\sqrt{2}}{141.42} \approx 1.06 \, A \]

### Step 7: Calculate the average power (P) dissipated across the resistor (R)
The average power (P) dissipated across the resistor is given by:
\[ P = I^2 R \]
Substituting the values:
\[ P = (1.06)^2 \times 100 \approx 112.36 \, W \]

### Step 8: Conclusion
- The average power dissipated across the resistor is approximately 112.36 W.
- The average power dissipated across the inductor and capacitor is 0 W because the phase difference between voltage and current in inductors and capacitors is 90 degrees.

### Summary of Results
1. Inductive Reactance (X_L) = 100 Ω
2. Capacitive Reactance (X_C) = 200 Ω
3. Impedance (Z) = 141.42 Ω
4. Average Power across Resistor (P) ≈ 112.36 W
5. Average Power across Inductor = 0 W
6. Average Power across Capacitor = 0 W