Quote:
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?
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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