Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

series LR

series RC

series LC

series LCR

Step-by-Step Solution

DC Behavior: When connected to a DC source, the frequency $\omega$ is zero. A capacitor offers infinite reactance ( $X_C = 1/\omega C = \infty$ ) to DC, blocking the current completely . Since a steady current of 0.4 A flows, the circuit cannot contain a capacitor in series. This eliminates options containing C (RC, LC, LCR).
Resistance: The resistance of the circuit can be calculated using the DC data: $R = V_{DC} / I_{DC} = 12 \text{ V} / 0.4 \text{ A} = 30 \, \Omega$ .
AC Behavior: When connected to an AC source, the circuit offers impedance $Z$ . Calculating impedance: $Z = V_{AC} / I_{AC} = 12 \text{ V} / 0.2 \text{ A} = 60 \, \Omega$ .
Conclusion: The impedance ( $60 \, \Omega$ ) is greater than the resistance ( $30 \, \Omega$ ). This indicates the presence of a reactive component ( $X$ ) such that $Z = \sqrt{R^2 + X^2}$ . Since capacitors are ruled out, the reactive component must be an inductor ( $X_L$ ). Thus, the circuit is a series combination of Inductance (L) and Resistance (R) .

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