AC Circuit Quiz #1

[Manoel]

Originally Posted by Ether

Suppose you have an air-core coil with inductance 200uH and resistance 0.09 ohms, connected in series with a 10 ohm resistor.

You apply a 15KHz 12 volt peak-to-peak sine wave across this series circuit.

What is the power dissipated in the coil?

What is the rms voltage measured across the coil?

Let's work exclusively with rms quantities: 12 Volts peak-to-peak is 6 Volts amplitude, and dividing by the square root of 2 (~ 1.41), we get an rms voltage of 4.24 V from the source.

At 15 kHz, the coil's reactance is given by:

XL = omega*L = 2*pi*f*L = 2*pi*15*10^3*200*10^-6 = 18.85 ohms

The coil's impedance, then

Zcoil = 0.09 + j*18.85

The circuit total impedance, considering the 10 ohm series resistor:

Ztotal = 10.09 + j*18.85

We can now determine the circuit's total current, in magnitude:

|I| = |V|/|Z| = |4.24|/|10.09+j*18.85| = 4.24/21.38 = 0.198 A

The power dissipated in the coil, then, is due to its small resistance:

P = i^2 * R = 0.198^2 * 0.09 = 0.0035 = 3.5 mW

We can determine the voltage drop in the coil using a voltage divider:

Vcoil = Vsource * Zcoil/Ztotal = 4.24*(0.09+j*18.85)/(10.09+j*18.85)
Vcoil = 3.3062+j*1.7495

The rms voltage that you would measure with a voltmeter is the magnitude of Vcoil:

|Vcoil| = |3.3062+j*1.7495|
|Vcoil| = 3.74 V