AC Circuit question problem

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  • Thread starter otester
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  • #1
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Was wondering if anyone can confirm that I am on the right track or not?

I'm not sure whether Irms is meant to be a complex number not :/

Thanks in advance,

otester

Homework Statement



A complex voltage v(t)=60sin(3wt)+15sin(3wt)+10sin(5wt) volts is applied to a coil of inductance 5mH and resistance 6(ohms). Determine, for a fudamental frequency of 100Hz, (a) an expression to represent the instantaneous value of current, (b) the rms voltage, (c) the rms current, (d) the power dissipated.


Homework Equations



XL = 2(pie)fL
Impedance (Z) = R + jXL
RMS current (Irms) = Vrms/Z
w = (omega)
Power dissipated=(Irms)x(Vrms)


The Attempt at a Solution



(a) XL = 2(pie)100x(5x10^-3) = 31.42(ohms)
Z = 6+j31.42
Irms = ( 60sin(3wt)+15sin(3wt)+10sin(5wt) ) / ( 6+j31.42 )

(b) Vrms = sqrt( (60^2+15^2+10^2)/2 ) = 44.3x0.707 = 31.32V

(c) Irms = 31.32 / ( 6+j31.42 ) = 5.22+j0.997
(In polar form: 5.31 @ 10.81 (degrees))

(d) Power dissipated = (31.32)x(5.31 @ 10.81) = 36.63 @ 10.81
 
Last edited:

Answers and Replies

  • #2
gneill
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A complete solution is offered.

I suspect that the first term of the given instantaneous voltage expression was meant to have an angular frequency of ω, not 3ω like the second term. Otherwise the two terms could simple by combined into a single term of 75 V magnitude (although this could be an intended trick for the student to catch). I will take the voltage expression to be:

v(t) = 60sin(ωt)+15sin(3ωt)+10sin(5ωt) volts

We are dealing with three frequencies and the circuit will exhibit three distinct impedances, one for each frequency component. We will use the superposition principle and deal with them separately.

##Z_1 = 6 + j 2\pi (f~L) = 6 + j \pi~Ω##
##Z_3 = 6 + j 2\pi (3f~L) = 6 + j3\pi ~Ω##
##Z_5 = 6 + j 2\pi (5f~L) = 6 + j5\pi~Ω##

Part (a) An expression for the instantaneous peak current
We divide each component of the peak voltage equation by the corresponding circuit impedance pertaining to its frequency. The results will be complex values. The magnitude of each result is the magnitude of the current term, while its phase angle is incorporated into the sin function of the time domain expression:

1) ##\frac{60}{6+j \pi} = 8.859~∠~-27.64°~~Amps##
2) ##\frac{15}{6 + j 2\pi} = 1.727~ ∠~-46.32°~~Amps##
3) ##\frac{10}{6 + j 5\pi} = 0.595~ ∠~-69.10°~~Amps##

##i(t) = 8.859sin(ωt - 27.64°) + 1.727sin(3ωt - 46.32°) + 0.595sin(5ωt - 69.10°)~~Amps##

Part (b) Determine the RMS voltage
The RMS values of the separate frequency components add in quadrature. So:

## V_{rms} = \sqrt{\left(\frac{60}{\sqrt{2}}\right)^2 + \left(\frac{15}{\sqrt{2}}\right)^2 + \left(\frac{10}{\sqrt{2}}\right)^2} = \sqrt{\frac{60^2 + 15^2 + 10^2}{2}} = 44.3~V##

Part (c) Determine the RMS current

Using the terms of the instantaneous peak current expression from part (a);
##I_{rms} = \sqrt{\frac{(8.859)^2 + (1.727)^2 + (0.595)^2}{2}} = 6.35~A##

Part(d) Determine the power dissipated
We'll sum the power dissipated at each frequency.

If V is an RMS voltage and I an RMS current and Φ the phase angle between current and voltage then the power is given by P = V*I cos(Φ). But if V and I are peak values, the √2 conversions to RMS of the voltage and current multiply, and the power becomes ½ V*I cos(Φ). We have all the peak voltage terms from the given information, and the peak currents and phases from part (a). So:

##
P = \frac{1}{2} [~~(60)(8.859) cos(-27.64°)\\
~~~~~~~~~~+ (15)(1.727) cos(-46.32°)\\
~~~~~~~~~~+ (10)(0.595) cos(-69.10°)~~] \\
~~= 241.9~W
##

So 242 W to three figures.
 
Last edited:
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@gneill you wrote a huge /Solution answer.Thank you so much!:smile:You are an awesome helper.
 

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