RLC Circuit (solved; but not sure where solution is incorrect)

[pious&peevish]

Homework Statement

A power supply with ΔVrms = 111 V is connected between points a and d in the figure below. (It's attached to the thread, but the basic idea is that this is an RLC circuit connected in series).

As shown, R = 39.5 Ω, L = 157 mH and C = 73.2 μF. At what frequency will it deliver a power of 210 W? Enter one of the two possible frequencies (in Hz).

Homework Equations

See Part 3 below

The Attempt at a Solution

We are given that P=I(rms)*V(rms), and by substituting in the values given in the problem we have 210 W = I(rms)*(111 V). Therefore our expected value for I(rms) should be 1.892 A.

I(rms)=V(rms)/Z, so 1.892 A = (111 V)/Z and the expected value for Z is 58.67 ohms.

58.67 ohms = √((39.5^2)+((2*pi*f*0.157 H)-(1/(2*pi*f*(73.2*(10^-6) F))^2).

From here, I solved for f, but the value I got was incorrect. I must have missed something somewhere, but I can't see what it is...

 

[SammyS]

Homework Statement

A power supply with ΔVrms = 111 V is connected between points a and d in the figure below. (It's attached to the thread, but the basic idea is that this is an RLC circuit connected in series).

As shown, R = 39.5 Ω, L = 157 mH and C = 73.2 μF. At what frequency will it deliver a power of 210 W? Enter one of the two possible frequencies (in Hz).

Homework Equations

See Part 3 below

The Attempt at a Solution

We are given that P=I(rms)*V(rms), and by substituting in the values given in the problem we have 210 W = I(rms)*(111 V). Therefore our expected value for I(rms) should be 1.892 A.

I(rms)=V(rms)/Z, so 1.892 A = (111 V)/Z and the expected value for Z is 58.67 ohms.

58.67 ohms = √((39.5^2)+((2*pi*f*0.157 H)-(1/(2*pi*f*(73.2*(10^-6) F))^2).

From here, I solved for f, but the value I got was incorrect. I must have missed something somewhere, but I can't see what it is...

P=I(rms)*V(rms) only if the the current and voltage (electric potential) are in phase.

The only element in an RLC circuit which dissipates energy is the resistor. Therefore, the average power dissipated by an RLC circuit is \displaystyle \ P_\text{Average}=\left(I_\text{RMS}\right)^2R\ .

 

[pious&peevish]

Thanks! So does that mean everything so far is right, except for the expression for power at the very beginning?

EDIT: Never mind -- I got the right answer; thanks!