Asked by Anonymous
What is the average power dissipation in “L” in an RLC series circuit with R = 10ohm, L = 0.1 H, C = 10 microF when driven by a 100 V-rms source?
Answers
Answered by
bobpursley
Wouldn't it depend on frequency?
The power dissipated is I^2*R
where I= 100Volts/ Z where Z is the resultant imedance.
The power dissipated is I^2*R
where I= 100Volts/ Z where Z is the resultant imedance.
Answered by
henry2,
At the resonant frequency:
Xl = -Xc
Xl+Xc = 0.
Therefore, Z = R = 116 ohms.
P = E^2/R = 100^2/116 = 86.2 W.
Xl = -Xc
Xl+Xc = 0.
Therefore, Z = R = 116 ohms.
P = E^2/R = 100^2/116 = 86.2 W.
Answered by
henry2,
Correction: Z = R = 10 ohms.
P = 100^2/10 = 1000 W.
P = 100^2/10 = 1000 W.
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