The variable resistor in the circuit in Fig. 10.24 is adjusted until maximum average power is delivered to $R_{L}$ .
a) What is the value of $R_{L}$ in ohms?
b) What is the maximum average power (in watts) delivered to $R_{L}$ ?

Question

The variable resistor in the circuit in Fig. 10.24 is adjusted until maximum average power is delivered to [latex]R_{L}[/latex].
a) What is the value of [latex]R_{L}[/latex] in ohms?
b) What is the maximum average power (in watts) delivered to [latex]R_{L}[/latex] ?

Accepted Answer

a) We first find the Thévenin equivalent with respect to the terminals of The circuit for determining the open circuit voltage in shown in Fig. 10.25. The variables[latex]\pmb{V}_{1},\pmb{V}_{2},\pmb{I}_{1},[/latex] and [latex] \pmb{I}_{2}[/latex] have been added to expedite the discussion.

First we note the ideal transformer imposes the following constraints on the variables [latex]\pmb{V}_{1},\pmb{V}_{2},\pmb{I}_{1},[/latex] and [latex] \pmb{I}_{2}[/latex] :

[latex]\pmb{V}_{2}=\frac{1}{4}\pmb{V}_{1}[/latex], [latex]\pmb{I}_{1}=-\frac{1}{4}\pmb{I}_{2}[/latex].

The open circuit value of [latex] \pmb{I}_{2}[/latex] is zero, hence [latex] \pmb{I}_{1}[/latex] is zero. It follows that

[latex]\pmb{V}_{1}=840 \angle 0° V,\pmb{V}_{2} = 210 \angle 0° V[/latex].

From Fig. 10.25 we note that [latex] \pmb{V}_{Th}[/latex] is the negative of [latex] \pmb{V}_{2}[/latex] , hence

[latex] \pmb{V}_{Th}= -210 \angle 0° V[/latex].

The circuit shown in Fig. 10.26 is used to determine the short circuit current. Viewing [latex] \pmb{I}_{1}[/latex] and [latex] \pmb{I}_{2}[/latex] as mesh currents, the two mesh equations are

[latex]840 \angle 0° = 80\pmb{I}_{1} - 20\pmb{I}_{2} + \pmb{V}_{1}[/latex],

[latex]0 = 20\pmb{I}_{2} - 20\pmb{I}_{1} + \pmb{V}_{2}[/latex].

When these two mesh current equations are combined with the constraint equations we get

[latex]840 \angle 0° = - 40\pmb{I}_{2} + \pmb{V}_{1}[/latex],

[latex]0 = 25\pmb{I}_{2} + \frac{\pmb{V}_{1}}{4}[/latex].

Solving for the short circuit value of [latex] \pmb{I}_{2}[/latex] yields

[latex] \pmb{I}_{2}=-6 A [/latex] .

Therefore the Thévenin resistance is

[latex]R_{Th} =\frac{-210}{-6}= 35 \Omega [/latex].

Maximum power will be delivered to [latex]R_{L}[/latex] when [latex]R_{L}[/latex]equals [latex] 35 \Omega [/latex].

b) The maximum power delivered to [latex]R_{L}[/latex] is most easily determined using the Thévenin equivalent. From the circuit shown in Fig. 10.27 we have

[latex]P_{max} =\left(\frac{-210}{70}\right)^{2}\left(35\right)= 315 W [/latex].

Question 10.11: The variable resistor in the circuit in Fig. 10.24 is adjust...

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