Question

A resistor of resistance 40 \( \Omega \), a capacitor of capacitive reactance 20 \( \Omega \) and an inductor of inductive reactance 50 \( \Omega \) are connected in series to an ac source of 100 V. The current through the circuit is:

• A. 0.5 A
• B. 1 A
• C. 1.5 A
• D. 2 A

Hint:

In a series AC circuit, always calculate the total impedance first and then apply Ohm's law to find the current. The impedance is the combined effect of resistance, capacitive reactance, and inductive reactance.

Answer 1

The Correct Option is D

Step 1: Calculate the total impedance \( Z \)
The total impedance in a series AC circuit with resistance \( R \), capacitive reactance \( X_C \), and inductive reactance \( X_L \) is given by: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Given:
\( R = 40 \, \Omega \)
\( X_C = 20 \, \Omega \)
\( X_L = 50 \, \Omega \)
Substitute the values into the formula: \[ Z = \sqrt{40^2 + (50 - 20)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \Omega \]
Step 2: Use Ohm's law to find the current
Now that we know the total impedance \( Z \), we can use Ohm's law to find the current \( I \): \[ I = \frac{V}{Z} \] Where \( V = 100 \, \text{V} \) (the voltage supplied by the source), and \( Z = 50 \, \Omega \) (the total impedance). Substitute the values: \[ I = \frac{100}{50} = 2 \, \text{A} \] Therefore, the current through the circuit is \( 2 \, \text{A} \).