Question 10.9: a) For the circuit shown in Fig. 10.23, what value of ZL res...

a) For the circuit shown in Fig. 10.23, what value of Z_{L} results in maximum average power transfer to Z_{L} ? What is the maximum power in milliwatts?
b) Assume that the load resistance can be varied between 0 and 4000Ω and that the capacitive reactance of the load can be varied between 0 and -2000 Ω. What settings of R_{L} and X_{L} transfer the most average power to the load? What is the maximum average power that can be transferred under these restrictions?

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a) If there are no restrictions on R_{L} and X_{L},the load impedance is set equal to the conjugate of the output or the Thévenin impedance. Therefore we set

R_{L}=3000Ω and X_{L}=-4000Ω,

or

Z_{L} = 3000 – j4000 Ω.

Because the source voltage is given in terms of its rms value, the average power delivered to Z_{L} is

P =\frac{1}{4}\frac{10^{2}}{3000} =\frac{25}{3} mW = 8.33 mW.

b) Because R_{L} and X_{L} are restricted, we first set X_{L} as close to -4000Ω as possible; thus X_{L}=-2000Ω. Next, we set R_{L} as close to \sqrt{R^{2}_{Th}+\left(X_{L}+X_{Th}\right)^{2}} as possible. Thus

R_{L} = \sqrt{3000^{2} + (-2000 + 4000)^{2}} = 3605.55 Ω.

Now, becauseR_{L} can be varied from 0 to 4000Ω, we can set R_{L} to 3605.55 Ω Therefore, the load impedance is adjusted to a value of

Z_{L}= 3605.55 – j2000 Ω.

With Z_{L} set at this value, the value of the load current is

I_{eff} =\frac{10 \angle 0°}{6605.55 + j2000} = 1.4489 \angle -16.85° mA.

The average power delivered to the load is

P = \left(1.4489 \times 10^{-3}\right)^{2}\left(3605.55\right) = 7.57 mW.

This quantity is the maximum power that we can deliver to a load, given the restrictions on R_{L} and X_{L}. Note that this is less than the power that can be delivered if there are no restrictions; in (a) we found that we can deliver 8.33 mW.

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