An alternating voltage source of 90\text{ V} (RMS) is connected in series across a circuit… — JAMB UTME

An alternating voltage source of $90\text{ V}$ (RMS) is connected in series across a circuit containing a resistor of resistance $12\ \Omega$ , an inductor of reactance $20\ \Omega$ , and a capacitor of reactance $11\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

A
$2.1\text{ A}$
B
$4.3\text{ A}$
$6.0\text{ A}$ Answer
D
$7.5\text{ A}$

Answer

The RMS current flowing through the circuit is

6.0\text{ A}

The total impedance

Z

of a series RLC circuit is given by

Z = \sqrt{R^2 + (X_L - X_C)^2}

. Substituting the given values gives

Z = \sqrt{12^2 + (20 - 11)^2} = \sqrt{144 + 81} = 15\ \Omega

. Dividing the RMS voltage (

90\text{ V}

) by this impedance yields an RMS current of

6.0\text{ A}

Step-by-Step Solution

Calculate the net reactance of the series AC circuit

X = X_L - X_C = 20\ \Omega - 11\ \Omega = 9\ \Omega

Inductive and capacitive reactances oppose each other in phase by

180^\circ

Determine the total impedance (Z) of the series RLC circuit using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\ \Omega

Resistance and net reactance are

90^\circ

out of phase, so impedance is calculated using the Pythagorean theorem.

Calculate the RMS current using Ohm's law for AC circuits

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{90\text{ V}}{15\ \Omega} = 6.0\text{ A}

The RMS current is equal to the RMS voltage divided by total circuit impedance.

Key Concept

Impedance and RMS Current in Series RLC Circuits

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