Question

A circuit when connected to an $AC$ source of $12\, V$ gives a current of $0.2\, A$. The same circuit when connected to a $DC$ source of $12\, V$, gives a current of $0.4 \,A$. The circuit is

• A. series LR
• B. series RC
• C. series LC
• D. series LCR

Answer 1

The Correct Option is A

To determine the type of circuit, we analyze the given information and apply relevant concepts of AC and DC circuits.

1. **Understanding the Circuit with AC Source:**

• When the circuit is connected to an AC source of 12\, V, it gives a current of 0.2\, A.
• The impedance (\(Z\)) of the circuit in an AC source is given by the formula:
  Z = \frac{V}{I} = \frac{12 \, V}{0.2 \, A} = 60\, \Omega.

2. **Understanding the Circuit with DC Source:**

• When the same circuit is connected to a DC source of 12\, V, it gives a current of 0.4 \,A.
• The resistance (\(R\)) of the circuit in a DC source is given by Ohm's Law:
  R = \frac{V}{I} = \frac{12 \, V}{0.4 \, A} = 30\, \Omega.

3. **Analysis of Circuit Components:**

• The difference between the values of impedance and resistance suggests the presence of inductive reactance (\(X_L\)) in the circuit.
• Impedance in an LR circuit is given by:
  Z = \sqrt{R^2 + X_L^2}.
• Substitute the known values:
  60 = \sqrt{30^2 + X_L^2}.
• Calculate \(X_L\):
  60^2 = 30^2 + X_L^2 \implies 3600 = 900 + X_L^2 \implies X_L^2 = 2700 \implies X_L = \sqrt{2700} \approx 51.96 \, \Omega.
• This calculation confirms the presence of inductive reactance, indicative of an LR series circuit.

4. **Conclusion:**

• Options such as RC or LC do not explain the observed phenomena as they would include capacitive reactance, which doesn't align with the given data.
• The current values and calculated impedance validly indicate the circuit is a series LR circuit.

Hence, the correct answer is series LR.