An alternating voltage source described by V(t) = 210\sqrt{2}\sin(100\pi t)\ \text{V} is c… — JAMB UTME

An alternating voltage source described by $V(t) = 210\sqrt{2}\sin(100\pi t)\ \text{V}$ is connected in series with a $40\ \Omega$ resistor, an inductor of inductive reactance $100\ \Omega$ , and a capacitor of capacitive reactance $70\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

$4.2\ \text{A}$ Answer
B
$1.0\ \text{A}$
C
$3.0\ \text{A}$
D
$7.0\ \text{A}$

Answer

The root-mean-square (RMS) current flowing through the circuit is

4.2\ \text{A}

The peak voltage is

V_0 = 210\sqrt{2}\ \text{V}

, giving an RMS voltage of

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

. The impedance of the series RLC circuit is calculated by

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = 50\ \Omega

. Therefore, the RMS current is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210}{50} = 4.2\ \text{A}

Step-by-Step Solution

Determine the RMS voltage from the voltage equation

V_{\text{rms}} = 210\ \text{V}

The standard equation for AC voltage is

V(t) = V_0 \sin(\omega t)

, where

V_0 = 210\sqrt{2}\ \text{V}

. The RMS voltage is

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

Calculate the total impedance of the series RLC circuit

Z = 50\ \Omega

Impedance is determined using phasor addition:

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\ \Omega

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

I_{\text{rms}} = 4.2\ \text{A}

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210\ \text{V}}{50\ \Omega} = 4.2\ \text{A}

Key Concept

Impedance and RMS current calculation in series RLC alternating current circuits

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