You have a 160- resistor and a 0.420-H inductor. Suppose you take the resistor a

Question

You have a 160- resistor and a 0.420-H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 34.0 V and an angular frequency of 240 rad/s .

A.What is the impedance of the circuit?

B.What is the current amplitude?

C.What is the voltage amplitude across the resistor?

D.What is the voltage amplitudes across the inductor?

E.What is the phase angle

of the source voltage with respect to the curren

F. Does the source voltage lag or lead the current

Explanation / Answer

The real part of the impedance (Z_r) is:

Z_r = 160

The imaginary part of the impedance (Z_i: is

Z_i = L

where = 240 rad/s and L = 0.420 H

Z_i = 240(0.420) =100.8

The magnitude of the impedance is the square root of the sum of the squares of the real and imaginary parts:

|Z| = sqrt{160² + 100.8²} 189.10

The magnitude of the current (I) is

|I| = 34.0 V/189.10 0.179 A

The magnitude (amplitude) of the voltage across the resistor is:

|V| = |I|Z_r = 0.179(160) = 28.64 V

The magnitude (amplitude) of the voltage across the inductor is:

|V| = |I|Z_i = 0.179(100.8) = 18.04 V

The phase angle () of the impedance is

= tan^-1(Z_i/Z_r) = tan^-1(100.8/160) 32.21°

To make the equation

V = I(Z)

come out that the voltage is at a zero net phase angle the current must be equal but opposite in sign with the impedance:

This makes the phase angle of the current become -32.21°

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