An alternating power supply of 220 V is applied across a series circuit of resistance `10sqrt(3)Omega,` capacitive reactance `40 Omega` and inductive reactance `30 Omega` The respective current in the circuit in the circuit for zero and infinite frequencies are

Options

A. `2 A, (1)/(2)A`
B. `0 A, 10 A`
C. `10 A, 0 A`
D. `0 A, 0 A`

Answer

To solve the problem, we need to find the current in a series LCR circuit for two specific cases: when the frequency (ω) is zero and when the frequency is infinite.

### Given:
- Voltage (V₀) = 220 V
- Resistance (R) = 10√3 Ω
- Capacitive Reactance (Xₐ) = 40 Ω
- Inductive Reactance (Xₗ) = 30 Ω

### Step 1: Calculate the Impedance (Z) for ω = 0
When the frequency is zero (ω = 0), the capacitive reactance (Xₐ) becomes infinite because Xₐ = 1/(ωC). Therefore, the circuit behaves as an open circuit.

- Impedance (Z) = √(R² + (Xₐ - Xₗ)²)
- At ω = 0, Xₐ = ∞, Xₗ = 30 Ω
- Z = √((10√3)² + (∞ - 30)²)
- Since (∞ - 30)² = ∞, we have Z = √(R² + ∞) = ∞

Now, we can calculate the current (I₀) using Ohm's law:
- I₀ = V₀ / Z
- I₀ = 220 V / ∞ = 0 A

### Step 2: Calculate the Impedance (Z) for ω = ∞
When the frequency is infinite (ω = ∞), the inductive reactance (Xₗ) becomes infinite because Xₗ = ωL. Therefore, the circuit behaves as an open circuit again.

- Impedance (Z) = √(R² + (Xₐ - Xₗ)²)
- At ω = ∞, Xₐ = 40 Ω, Xₗ = ∞
- Z = √((10√3)² + (40 - ∞)²)
- Since (40 - ∞)² = ∞, we have Z = √(R² + ∞) = ∞

Now, we can calculate the current (I₀) again:
- I₀ = V₀ / Z
- I₀ = 220 V / ∞ = 0 A

### Final Results:
- Current for ω = 0: I₀ = 0 A
- Current for ω = ∞: I₀ = 0 A

### Conclusion:
The respective current in the circuit for zero and infinite frequencies is 0 A for both cases.