Question

From the given A.C. circuit, find out:
Inductive and capacitive reactance
Frequency of the applied voltage in the state of resonance
Impedance of the circuit in resonance stage

Hint:

At resonance, $X_L = X_C$ and the circuit behaves like a pure resistor.

Answer 1

Step 1: Identify given values.
\[ L = 20 \, \text{mH} = 20 \times 10^{-3} \, H, \quad C = 2 \, \mu F = 2 \times 10^{-6} \, F, \quad R = 1000 \, \Omega. \]
Step 2: Inductive reactance.
\[ X_L = 2 \pi f L. \]
Step 3: Capacitive reactance.
\[ X_C = \frac{1}{2 \pi f C}. \]
Step 4: Resonance condition.
At resonance, $X_L = X_C$. So, \[ 2 \pi f L = \frac{1}{2 \pi f C}. \] \[ f^2 = \frac{1}{(2 \pi)^2 L C}. \] \[ f = \frac{1}{2 \pi \sqrt{LC}}. \]
Step 5: Substitution.
\[ f = \frac{1}{2 \pi \sqrt{20 \times 10^{-3} \times 2 \times 10^{-6}}}. \] \[ f = \frac{1}{2 \pi \sqrt{40 \times 10^{-9}}}. \] \[ f = \frac{1}{2 \pi \times 2 \times 10^{-4.5}} \approx 1000 \, \text{Hz}. \]
Step 6: Reactances at resonance.
\[ X_L = 2 \pi f L = 2 \pi (1000)(20 \times 10^{-3}) = 125.6 \, \Omega. \] \[ X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (1000)(2 \times 10^{-6})} = 125.6 \, \Omega. \]
Step 7: Impedance at resonance.
At resonance, $X_L = X_C$, so impedance is purely resistive: \[ Z = R = 1000 \, \Omega. \]
Step 8: Conclusion.
(A) $X_L = 125.6 \, \Omega$, $X_C = 125.6 \, \Omega$
(B) $f = 1000 \, \text{Hz}$
(C) $Z = 1000 \, \Omega$