You have a 210 ohm resistor, a 0.400 H inductor, and a 6.00 mu F capacitor. Supp

Question

You have a 210 ohm resistor, a 0.400 H inductor, and a 6.00 mu F 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. What is the impedance of the circuit? ohm What is the current amplitude? A What are the voltage amplitudes across the resistor and across the inductor? V (across the resistor) V (across the inductor) What is the phase angle phi of the source voltage with respect to the current? degree

Explanation / Answer

The real part of the impedance (Z_r) is:

Z_r = 210

The imaginary part of the impedance (Z_i: is

Z_i = L

where = 250 rad/s and L = 0.400 H

Z_i = 250x(0.400) = 100

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

|Z| = sqrt{210² + 100²} 232.59

The magnitude of the current (I) is

|I| = 30.0 V/232.59 = 0.1289 A

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

|V| = |I|Z_r = 0.1289x(210) = 27.069 V

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

|V| = |I|Z_i = 0.1289x(100) = 12.89 V

The phase angle () of the impedance is

= tan^-1(Z_i/Z_r) = tan^-1(100/210) = 25.46°

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 -25.46°

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