Question

A 200Ω resistor, 0.900H inductor, and 6.00µF capacitor are connected in series across a voltage source that has voltage amplitude 30.0 V and an angular frequency of 250 rad/s. (a) What are v, v_R, v_L, and v_C at t = 20.0 ms? Compare v_R + v_L + v_C to v at this instant. (b) What are V_R, V_L, and V_C? Compare V to V_R + V_L + V_C. Explain why these two quantities are not equal.

Short answer

a. On comparing the instantaneous supply voltage to the summation of instantaneous voltage across individual electric components, we can conclude that both of them are equal.

$V_R+V_L+V_c=V$

b. On comparing the amplitude of voltage of the source to the summation of amplitude of voltage across individual electric components, we can conclude that both of them are not equal

$V_R+V_L+V\ne V$

Step-by-step solution

Concept

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

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).

Given values

Resistance of the resistor, R = 200 Ω

The inductance of the coil, L = 0.900 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

role="math" localid="1668251678662" $\begin{array}{r}{X}_L=\omega L\\ =(250\mathrm{rad}/\mathrm{s})\left(0.9\mathrm{H}\right)\\ =225\mathrm{\Omega }\end{array}$

Putting the values in Z

$\begin{array}{r}Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}\\ =\sqrt{{200}^2+(225-666.67{)}^2}\\ =484.84\mathrm{\Omega }\end{array}$

Therefore, the impedance of the circuit is 484.84 Ω

Determination of Current Amplitude

By using Ohm’s Law

$\begin{array}{r}V=I.Z\\ I=\frac{V}{Z}\end{array}$

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

$I=\frac{V}{Z}=\frac{30.0V}{484.84\Omega }=0.0619A$

Therefore, the amplitude of the current in circuit is 0.0619 A

Determination of Phase angle

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

$\begin{array}{r}\tan \varphi =\frac{\left(X_L-X_C\right)}{R}\\ \tan \varphi =\frac{(225-666.67)\mathrm{\Omega }}{200\mathrm{\Omega }}\\ \tan \varphi =-2.2084\\ \varphi =-{65.64}^{\circ }\end{array}$

Therefore, the phase angle of the voltage source with respect to current is -65.64° which means voltage lags the current as the phase angle is negative.

Determination of Voltage amplitude and Instantaneous Voltages

The instantaneous voltage across resistor:

$\begin{array}{r}{V}_R=IR\\ =\left(0.0619A\right)\left(200\mathrm{\Omega }\right)\\ =12.38\mathrm{V}\end{array}$

As the voltage across resistor is in phase with current, it can be represented as:

$v_R=V_{R}cos\omega t=12.38cos\left[\right(250rad/s\left)t\right]$

The instantaneous voltage across inductor:

$\begin{array}{r}{V}_L=I{X}_L\\ =\left(0.0619\mathrm{A}\right)\left(225\mathrm{\Omega }\right)\\ =13.928\mathrm{V}\end{array}$

As the voltage across inductor leads the current, it can be represented as:

$v_L=-V_{L}sin\omega t=-13.928sin\left[\right(250rad/s\left)t\right]$

Voltage drop across capacitor is given by

$\begin{array}{r}{V}_C=L{X}_{\mathrm{C}}\\ =\left(0.0619A\right)\left(666.67\mathrm{\Omega }\right)\\ =41.267\mathrm{V}\end{array}$

As the voltage across capacitor lags the current, it can be represented as:

$v_C=V_{C}sin\omega t=41.267sin\left[\right(250rad/s\left)t\right]$

Thus, the summation of all the value gives at t = 20 ms can be given as:

$\begin{array}{r}{v}_R+v_L+v_C=12.38\cos \left[(250\mathrm{rad}/\mathrm{s})\left({20}^{\ast }{10}^{-3}\right)\right]-13.928\sin \left[(250\mathrm{rad}/\mathrm{s})\left(20\ast {10}^{-3}\right)\right]\\ +41.267\sin \left[(250\mathrm{rad}/\mathrm{s})\left({20}^{\ast }{10}^{-3}\right)\right]\\ =3.51+13.35-39.55\\ =-22.70\mathrm{V}\end{array}$

Therefore, the net instantaneous voltage across resistor, inductor, and capacitor is -22.70 volts.

Comparison of Instantaneous supply voltage

The instantaneous supply voltage at t = 20 ms can be given as:

$\begin{array}{r}v=V\cos [\omega t+\varphi ]\\ =30\cos \left[250\ast 20\ast {10}^{-3}-65.64\ast \frac{\pi }{180}\right]\\ =-22.7\mathrm{V}\end{array}$

From step 4 and 5 we can conclude that

$v_R+v_L+v_C=v$

Therefore, on comparing the instantaneous supply voltage to the summation of instantaneous voltage across individual electric components, we can conclude that both of them are equal.

Comparison of Voltage amplitudes

From step 5, amplitudes of voltage across each electric component can be listed as:

$\begin{array}{l}{V}_R=12.4V\\ V_L=139V\\ V_c=413V\end{array}$

The summation of all these voltages:

$$\begin{gathered} V_R+V_L+V_c=124+139+41.3 \\ =67.6V \end{gathered}$$

Thus, we can conclude that

$$\begin{gathered} V_R+V_t+V\ne V \\ 67.6V\ne 30V \end{gathered}$$

Therefore, on comparing the amplitude of voltage of the source to the summation of amplitude of voltage across individual electric components, we can conclude that both of them are not equal.