Question

The resistor, inductor, capacitor, and voltage source described in Exercise 31.14 are connected to form an L-R-C series circuit. (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? (d) What are the voltage amplitudes across the resistor, inductor, and capacitor? (e) Explain how it is possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across the source.

Short answer

1. The impedance of the circuit is 600 Ω
2. The amplitude of the current in circuit is 0.05 A
3. The phase angle of the voltage source with respect to current is -70.56° which means voltage lags the current as the phase angle is negative.
4. The voltage drop across the resistor, inductor, and capacitor is 10 V, 5 V, and 33.33 V respectively.
5. The voltage drop across capacitor could get larger than voltage of source when net sum of is$v_L+v_R$ negative.

Step-by-step solution

Concept

An inductor is a passive two-terminal device that stores energy in a magnetic field when current passes through it. When an inductor is attached to an AC supply, the resistance produced by it is called inductive reactance (X_L).

A capacitor is a device consisting of two conductors in close proximity that are used to store electrical energy. These conductors are insulated from each other. When a capacitor is attached to an AC supply, the resistance produced by it is called capacitive reactance (X_C).

Resistance is measure of opposition to the flow of current in a closed electrical circuit. It is measured in Ohm (Ω).

Impedance is defined as the effective resistance of an electric circuit to the flow of current due to the combined effect of resistance (offered by resistor) and reactance (offered by capacitor and inductor).

Given values

Resistance of the resistor, R = 200 Ω

The inductance of the coil, L = 0.400 H

Capacitance of capacitor, C = 6.00 µF

The amplitude of voltage source, V = 30.0 V

Angular frequency of the source, ω = 250 rad/s

Determination of Circuit’s Impedance

Since the capacitor, inductor, and resistor are connected in series, Impedance of the circuit will be given by

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

Here, X_C is the inductive reactance that is given by

$\begin{array}{r}{X}_c=\frac{1}{\omega C}\\ =\frac{1}{(250\mathrm{rad}/\mathrm{s})\left(6.00\ast {10}^{-6}\mathrm{F}\right)}\\ =666.67\mathrm{\Omega }\end{array}$

And, X_L is the inductive reactance that is given by

$\begin{array}{r}{X}_L=\omega L\\ =(250\mathrm{rad}/\mathrm{s})\left(0.4\mathrm{H}\right)\\ =100\mathrm{\Omega }\end{array}$

Putting the values in Z

$\begin{array}{r}Z=\sqrt{R^2+{\left(X_C-X_L\right)}^2}\\ =\sqrt{{200}^2+(666.67-100{)}^2}\\ =600\mathrm{\Omega }\end{array}$

Therefore, the impedance of the circuit is 600 Ω

Determination of Current Amplitude

By using Ohm’s Law

$\begin{array}{l}V=IZ\\ I=\frac{V}{Z}\end{array}$

It is given that voltage amplitude of the source is 30.0 V, so

role="math" localid="1668251293354" $I=\frac{V}{Z}=\frac{300V}{601\Omega }=0.05A$

Therefore, the amplitude of the current in circuit is 0.05 A

Determination of Phase angle

The angle between voltage and current phasors in L-C- circuit is given by

$\begin{array}{r}\tan \varphi =\frac{\left(X_L-X_C\right)}{R}\\ \tan \varphi =\frac{(100-666.67)\mathrm{\Omega }}{200\mathrm{\Omega }}\\ \tan \varphi =-2.833\\ \varphi =-{70.56}^{\circ }\end{array}$

Therefore, the phase angle of the voltage source with respect to current is -70.56° which means voltage lags the current as the phase angle is negative.

Determination of Voltage across R, L, and C

Voltage drop across the resistor is in phase with current, so by using ohm’s law

$\begin{array}{r}{V}_R=IR\\ =\left(0.05A\right)\left(200\mathrm{\Omega }\right)\\ =10\mathrm{V}\end{array}$

Voltage drop across inductor is given by

$\begin{array}{r}{V}_L=I{X}_L\\ =\left(0.05\mathrm{A}\right)\left(100\mathrm{\Omega }\right)\\ =5\mathrm{V}\end{array}$

Voltage drop across capacitor is given by

$\begin{array}{r}{V}_C=I{X}_{\mathrm{C}}\\ =\left(0.05A\right)\left(666.67\mathrm{\Omega }\right)\\ =33.33\mathrm{V}\end{array}$

Therefore, the voltage drop across the resistor, inductor, and capacitor is 10 V, 5 V, and 33.33 V respectively.

Voltage across capacitor

The voltage drop across capacitor could get larger than voltage of source when net sum of $v_L+v_R$is negative.