An LCR series circuit consisting of resistance \( 45\,\Omega \), inductive reactance \( 4\,\Omega \) and capacitive reactance \( 4\,\Omega \) is connected to an alternating source of emf 90 V. The readings of the ammeter connected to the circuit would be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 A
The current in a series LCR circuit is the voltage divided by the impedance, and the impedance depends on the difference of the two reactances.
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \)
Here \( X_L = X_C = 4\,\Omega \), so the reactances cancel and the circuit is at resonance. The impedance is just the resistance.
\( \displaystyle Z = \sqrt{45^{2} + 0} = 45\,\Omega \)
An ac voltage quoted without comment is its rms value, and an ac ammeter reads the rms current.
\( \displaystyle I = \frac{V}{Z} = \frac{90\,\mathrm{V}}{45\,\Omega} = 2\,\mathrm{A} \)