Refer to Fig. 12.53. Express the average power dissipated in the load RL as a function of RL and the other circuit parameters. Let the frequency of the source be f=30kHz and draw a graph of the power dissipated in the load versus the load resistance RL, for 1Ω≤RL≤10kΩ. Use a logarithmic scale for

Question

Refer to Fig. 12.53. Express the average power dissipated in the load [latex]R_L[/latex] as a function of [latex]R_L[/latex] and the other circuit parameters. Let the frequency of the source be [latex]f=30 \mathrm{kHz}[/latex] and draw a graph of the power dissipated in the load versus the load resistance [latex]R_L[/latex], for [latex]1 \Omega \leq R_L \leq 10 \mathrm{k} \Omega[/latex]. Use a logarithmic scale for the resistance. From the graph, determined (or estimate) the load resistance for which the power dissipated has its maximum value. Then let the load have that resistance and draw a graph of the power dissipated in the load versus the frequency [latex]f[/latex] of the source, for [latex]100 \mathrm{~Hz} \leq f \leq 100 \mathrm{MHz}[/latex] and estimate the frequency for which the power dissipated in the load is maximum.

Accepted Answer

We first obtain the Thévenin equivalent for the circuit to the left of the load [latex]R_L[/latex]. Kirchhoff's current law gives

[latex] \begin{aligned} & ilde{V}_T\left(\frac{1}{R_S}+\frac{1}{R+j \omega L}+j \omega C ight)=\frac{V_S}{R_S} \\ &\Rightarrow ilde{V}_T=\frac{(R+j \omega L) V_S}{R_S+R-R_S L C \omega^2+j \omega\left(L+R_S R C ight)} \end{aligned} [/latex]

where [latex] u_S[/latex] is the phase reference and [latex] ilde{V}_T[/latex] is the open-circuit (Thévenin equivalent) voltage. The short-circuit (Norton equivalent) current is given by

[latex] ilde{I}_N=\frac{V_S}{R_S}, [/latex]

so the Thévenin equivalent impedance is given by

[latex] Z_T=\frac{ ilde{V}_T}{ ilde{I}_N}=\frac{(R+j \omega L) R_S}{R_S+R-R_S L C \omega^2+j \omega\left(L+R_S R C ight)}, [/latex]

where f is the frequency of the source and [latex]\omega=2 \pi f [/latex]. The rms amplitude of the voltage across a load [latex]R_L [/latex] is given by

[latex] V_{L \mathrm{rms}}=\left|\frac{V_S R_L}{\sqrt{2}\left(Z_T+R_L ight)} ight| [/latex]

and the power dissipated by the load is given by

[latex] P_L=\frac{V_{L \mathrm{rms}^2}}{R_L} . [/latex]

Figure 12.54 shows a (computer-generated) graph of the power dissipated versus the load resistance. The load resistance that draws maximum power is approximately [latex]R_L=10 \Omega [/latex]. To see if this result is reasonable, we calculate the Thévenin impedance for [latex]f=30 \mathrm{kHz} [/latex]. We obtain [latex]Z_T=(9.35+j 2.18) \Omega [/latex], so we might expect maximum power transfer for a load in the neighborhood of [latex]10 \Omega [/latex]. We also expect the power dissipated to approach zero for [latex]R_L ightarrow 0 [/latex], in which case the load voltage approaches zero, and for [latex]R_L ightarrow \infty [/latex], in which case the load current approaches zero. The graph in Fig. 12.54 is consistent with these expectations.

Figure 12.55 shows a computer-generated graph of the power dissipated in the load [latex]R_L=10 \Omega [/latex] versus frequency. Maximum power is dissipated for frequencies near [latex] 100 \mathrm{kHz} [/latex]. This is not surprising, because the magnitude of the impedance of the parallel RLC branch at that frequency is [latex]678 \Omega [/latex], which is much larger than both the source resistance and the load resistance. Thus the parallel RLC branch draws very little current for frequencies near [latex]100 \mathrm{kHz} [/latex]. For frequencies approaching zero, the impedance of the parallel RLC branch approaches [latex]R=2 \Omega [/latex], so the load voltage approaches

[latex] V_{L \mathrm{rms}}(f ightarrow 0)=\frac{R \| R_L}{R_S+R \| R_L} \frac{V_S}{\sqrt{2}} \cong 1.52 \mathrm{~V}, [/latex]

and the power dissipated approaches

[latex] P_L(f ightarrow 0)=\frac{V_{L \mathrm{rms}}^2}{R_L} \cong 230 \mathrm{~mW} . [/latex]

For frequencies much larger than [latex]100 \mathrm{kHz} [/latex], the impedance of the parallel RLC branch approaches the impedance of the capacitor,

which in turn approaches zero, so the parallel RLC branch shunts virtually all of the current to ground and the power dissipated by the load approaches zero. The graph in Fig. 12.55 is consistent with these observations.

This example illustrates using a Thévenin equivalent to simplify studying load power (or current or voltage) as a function of the load impedance and source frequency. Using the Thévenin equivalent means we must solve equations obtained from (e.g.) Kirchhoff's current law only once. Thereafter, the circuit is simply a voltage divider, and subsequent analysis is relatively simple.

Question 12.41: Refer to Fig. 12.53. Express the average power dissipated in...

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