Question

You have a 200Ω resistor, a 0.400-H inductor, and a 6.00µF capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 V and an angular frequency of 250 rad/s. (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What are the voltage amplitudes across the resistor and across the inductor? (d) What is the phase angle ϕ of the source voltage with respect to the current? Does the source voltage lag or lead the current? (e) Construct the phasor diagram.

Short answer

a) The impedance of the circuit is 224 Ω

b) The amplitude of the current in circuit is 0.134 A

c) The voltage drop across the resistor and inductor is 26.8 V and 13.4 V respectively

d) The phase angle of the voltage source with respect to current is +26.6°

e) The phasor diagram representing the magnitude and directional relationship between current and voltage in the circuit is given by

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).

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).

A phasor diagram is graphical representation of magnitude and directional relationship between two or more alternating quantities.

Given information

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

$Z=\sqrt{R^2+X_L^2}$

Here, XL is the inductive reactance and its value is given by

$$\begin{gathered} X_L=\omega L \\ =\left(250rad/s\right)\left(0.4H\right) \\ =100\Omega \end{gathered}$$

Putting the values in Z

$$\begin{gathered} Z=\sqrt{R^2+X_L^2} \\ =\sqrt{{200}^2+{100}^2} \\ =224\Omega \end{gathered}$$

Therefore, the impedance of the circuit is 224 Ω

Determination of Current Amplitude

By using Ohm’s Law

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

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

$I=\frac{V}{Z}=\frac{30.0V}{224\Omega }=0.134A$

Therefore, the amplitude of the current in circuit is 0.134 A

Determination of Voltage across Resistor and Inductor

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

$$\begin{gathered} V_R=IR \\ =\left(0.134A\right)\left(200\Omega \right) \\ =26.8V \end{gathered}$$

The voltage drop across the inductor is given by

$$\begin{gathered} V_L=I{X}_L \\ =\left(0.134A\right)\left(100\Omega \right) \\ =13.4V \end{gathered}$$

Therefore, the voltage drop across the resistor and inductor is 26.8 V and 13.4 V respectively.

Determination of Phase angle

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

$$\begin{gathered} \tan \varphi =\frac{X_L}{R} \\ \tan \varphi =\frac{100\Omega }{200\Omega } \\ \tan \varphi =0.5 \\ \varphi =+26.6° \end{gathered}$$

Therefore, the phase angle of the voltage source with respect to current is +26.6°

Phasor Diagram

The phasor diagram representing the magnitude and directional relationship between current and voltage in the circuit is given by