The circuit to the left of terminals A and B in Figure 19-39 provides power to the load ZL. This can be viewed as simulating a power amplifier delivering power to a complex load.It is the Thevenin equivalent of a more complex circuit.Calculate and plot a graph of the power delivered to the load for

Question

The circuit to the left of terminals A and B in Figure 19-39 provides power to the load [latex] Z_{L}[/latex]. This can be viewed as simulating a power amplifier delivering power to a complex load .It is the Thevenin equivalent of a more complex circuit. Calculate and plot a graph of the power delivered to the load for each of the following frequencies: 10 kHz, 30 kHz. 50 kHz, 80 kHz, and 100 kHz.

Accepted Answer

For f= 10 kHz,

[latex]X_{C}= \frac{1}{2\pi fC}= \frac{1}{2\pi (10 \ kHz)(0.01 \ \mu F)}= 1.59 \ k\Omega [/latex]

[latex]X_{L}= 2\pi fL=2\pi (10 \ kHz)(1 \ mH)= 62.8 \ \Omega[/latex]

The magnitude of the total impedance is

[latex]Z_{tot}= \sqrt{(R_{s}+ R_{L})^{2}+ (X_{L}- X_{C})^{2}} = \sqrt{(20 \ \Omega )^{2}+ (1.53 \ k\Omega )^{2}}= 1.53 \ k\Omega[/latex]

The current is

[latex]I= \frac{V_{s}}{Z_{tot}}= \frac{10 \ V}{1.53 \ k\Omega }= 6.54 \ mA[/latex]

The load power is

[latex]P_{L}= I^{2}R_{L}= (6.54 \ mA)^{2}(10 \ \Omega )=428 \ \mu W[/latex]

For f= 30 kHz,

[latex]X_{C}= \frac{1}{2\pi (30 \ kHz)(0.01 \ \mu F)}= 531 \ \Omega [/latex]

[latex]X_{L}=2\pi (30 \ kHz)(1 \ mH)= 189 \ \Omega [/latex]

[latex]Z_{tot}= \sqrt{(20 \ \Omega )^{2}+ (342 \ \Omega )^{2}}= 343 \ \Omega[/latex]

[latex]I= \frac{V_{s}}{Z_{tot}}= \frac{10 \ V}{343 \ \Omega }= 29.2 \ mA[/latex]

[latex]P_{L}= I^{2}R_{L}= (29.2 \ mA)^{2}(10 \ \Omega )=8.53 \ mW[/latex]

For f= 50 kHz,

[latex]X_{C}= \frac{1}{2\pi (50 \ kHz)(0.01 \ \mu F)}= 318 \ \Omega [/latex]

[latex]X_{L}=2\pi (50 \ kHz)(1 \ mH)= 314 \ \Omega [/latex]

Note that [latex]X_{C}[/latex] and [latex]X_{L}[/latex] are very close to being equal which makes the impedances approximately complex conjugates. The exact frequency at which [latex]X_{L}[/latex] = [latex]X_{C}[/latex] is 50.3 kHz.

[latex]Z_{tot}= \sqrt{(20 \ \Omega )^{2}+ (4 \ \Omega )^{2}}= 20.4 \ \Omega[/latex]

[latex]I= \frac{V_{s}}{Z_{tot}}= \frac{10 \ V}{20.4 \ \Omega }= 490 \ mA[/latex]

[latex]P_{L}= I^{2}R_{L}= (490 \ mA)^{2}(10 \ \Omega )=2.40 \ W[/latex]

For f= 80 kHz,

[latex]X_{C}= \frac{1}{2\pi (80 \ kHz)(0.01 \ \mu F)}= 199 \ \Omega [/latex]

[latex]X_{L}= 2\pi (80 \ kHz)(1 \ mH)= 503 \ \Omega [/latex]

[latex]Z_{tot}= \sqrt{(20 \ \Omega )^{2}+ (304 \ \Omega )^{2}}= 305 \ \Omega[/latex]

[latex]I= \frac{V_{s}}{Z_{tot}}= \frac{10 \ V}{305 \ \Omega }= 32.8 \ mA[/latex]

[latex]P_{L}= I^{2}R_{L}= (32.8 \ mA)^{2}(10 \ \Omega )=10.8 \ mW[/latex]

For f= 100 kHz.

[latex]X_{C}= \frac{1}{2\pi (100 \ kHz)(0.01 \ \mu F)}= 159 \ \Omega [/latex]

[latex]X_{L}= 2\pi (100 \ kHz)(1 \ mH)= 628 \ \Omega [/latex]

[latex]Z_{tot}= \sqrt{(20 \ \Omega )^{2}+ (469 \ \Omega )^{2}}= 469 \ \Omega[/latex]

[latex]I= \frac{V_{s}}{Z_{tot}}= \frac{10 \ V}{469 \ \Omega }=21.3 \ mA[/latex]

[latex]P_{L}= I^{2}R_{L}= (21.3 \ mA)^{2}(10 \ \Omega )=4.54 \ mW[/latex]

As you can see from the results, the power to the load peaks at the frequency (50 kHz) for which the load impedance is the complex conjugate of the output impedance (when the reactances are equal in magnitude). A graph of the load power versus frequency is shown in Figure 19-40. Since the maximum power is so much larger than the other values, an accurate plot is difficult to achieve without intermediate values.

Question 19.16: The circuit to the left of terminals A and B in Figure 19-39...

Related Answered Questions