How Is Power Transferred to a Load in a Circuit with a Cosine Voltage Source?

[Bromio]

Homework Statement

Calculate the power transferred when Vi(t) = 50 cos (4π 10t) in the attached figure.

Homework Equations

P(load) = |V(load rms)||I(load rms)|cos(phase(V)-phase(I))

The solution must be P(load) = 3.82 W.

The Attempt at a Solution

I've calculated the Thevenin equivalent of the circuit outside the load. Vth = 27.1<12.5º and Zth = 12.1<75.96º.

I've also calculated Zload = 50 Ω approximately.

V(load rms) is obtained by the voltage divisor. I got about 17.7<0º and I(load rms) = V(load rms)/Z(load) = 0.353<0º.

So, using the formula in section 2., P(load) = 6.24 W, which is not the correct result.

Where have I made a mistake?

Thank you.

 

[gneill]

The given solution of 3.82 W doesn't look credible for the given circuit. The Thevenin equivalent values that were obtained also don't make much sense, but since the OP did not supply details on the work done to obtain them I can't comment on what went wrong. I can however look into what the correct answer should be.

The operating frequency of the voltage source is pretty low: 20 Hz, so those tiny capacitors are going to have very high impedances compared to the circuit's resistors. The inductor is also small (only 0.318 μH) and at 20 Hz its impedance will be very small. Let's check:

$ω = 4 \pi 10 = 125.66~rad/sec$

$Z_{C1} = \frac{1}{j ω 536~pF} = -j~1.485×10^7~Ω$

$Z_{C2} = \frac{1}{j ω 100~pF} = -j~7.958×10^7~Ω$

$Z_L = j ω 0.318 μH = j~3.996×10^{-5}~Ω$

(Extra digits are kept for these values since they may be used in further calculations and we don't want rounding errors to intrude on our significant figures) We can see that these components really aren't going to affect the power transfer to the 50 Ohm load significantly. A very good approximation of the load voltage and current can thus be obtained by ignoring the reactive components and considering the resistors only. Let's do that.

The way that the source function is specified we can assume that the 50 V magnitude is a peak value. We should convert this to an RMS value if we're looking for power. So, let $E = 50/\sqrt{2} = 35.36~V$. Then the load resistor current and voltage are:

$I_L = \frac{35.36~V}{60~Ω} = 0.5893~A$ (RMS)

$V_L = I×50~Ω = 29.46~V$ (RMS)

and the power will be:

$P = V I = 17.36~W$

So the "true" answer should be very close to 17.4 watts.