Question

In the series L-C-R circuit shown, the impedance is

• A. 200 \( \Omega \)
• B. 100 \( \Omega \)
• C. 300 \( \Omega \)
• D. 500 \( \Omega \)

Hint:

To calculate the impedance in a series L-C-R circuit, use the formula \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), and remember that \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance.

Answer 1

The Correct Option is D

To solve the problem, we need to calculate the impedance of an RLC circuit with given component values.

1. Given Parameters:
- Inductance, $L = 1 \, \text{H}$
- Capacitance, $C = 20 \, \mu \text{F} = 20 \times 10^{-6} \, \text{F}$
- Resistance, $R = 300 \, \Omega$
- Frequency, $f = \frac{50}{\pi} \, \text{Hz}$

2. Calculate Inductive Reactance ($X_L$):
The inductive reactance is given by:

$ X_L = 2 \pi f L $
Substituting the values:

$ X_L = 2 \pi \left( \frac{50}{\pi} \right) \times 1 = 100 \, \Omega $

3. Calculate Capacitive Reactance ($X_C$):
The capacitive reactance is given by:

$ X_C = \frac{1}{2 \pi f C} $
Substituting the values:

$ X_C = \frac{1}{2 \pi \left( \frac{50}{\pi} \right) (20 \times 10^{-6})} = 500 \, \Omega $

4. Calculate Total Impedance ($Z$):
The impedance of an RLC circuit is:

$ Z = \sqrt{R^2 + (X_C - X_L)^2} $
Substituting the values:

$ Z = \sqrt{(300)^2 + (500 - 100)^2} = \sqrt{90000 + 160000} = 500 \, \Omega $

Final Answer:
The impedance of the circuit is $500 \, \Omega$.

Answer 2

To solve the problem, we need to calculate the impedance of a series L-C-R circuit.

1. Calculate the Inductive Reactance ($X_L$):
The inductive reactance is given by:

$ X_L = \omega L = 2\pi f L $
where $f$ is the frequency and $L$ is the inductance. Given $L = 1 \, \text{H}$ and $f = \frac{50}{\pi} \, \text{Hz}$,

$ X_L = 2\pi \left(\frac{50}{\pi}\right) (1) = 100 \, \Omega $

2. Calculate the Capacitive Reactance ($X_C$):
The capacitive reactance is given by:

$ X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} $
where $C = 20 \times 10^{-6} \, \text{F}$ (20 µF). Substituting the given values:

$ X_C = \frac{1}{2\pi \left(\frac{50}{\pi}\right) (20 \times 10^{-6})} = \frac{1}{100 \times 20 \times 10^{-6}} = \frac{1}{0.002} = 500 \, \Omega $

3. Calculate the Impedance ($Z$):
In a series L-C-R circuit, the impedance is given by:

$ Z = \sqrt{R^2 + (X_L - X_C)^2} $
where $R = 300 \, \Omega$. Substituting the values:

$ Z = \sqrt{(300)^2 + (100 - 500)^2} = \sqrt{(300)^2 + (-400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega $

Final Answer:
The impedance of the series L-C-R circuit is $ {500 \, \Omega} $.