In the network shown in Figure 5.105, determine the value of load resistance RL to give maxi-mum power transfer and the power delivered to the load.

Question

In the network shown in Figure 5.105, determine the value of load resistance [latex]R_{L}[/latex] to give maxi-mum power transfer and the power delivered to the load.

Accepted Answer

The equivalent resistance of the network when viewed from terminals A and B after removing resistor [latex]R_{L}[/latex] from the circuit and short circuiting the voltage source (see Figure 5.106(a)), we get

[latex] R_{th} = 4 + \left(10 \parallel 10 ight) = 4 + \frac{10 imes 10}{10 + 10} = 9 \Omega[/latex]

Maximum power will be transferred to the load, when it is equal to Thevenin’s equivalent resistance [latex]R_{th}[/latex],

[latex] R_{L} = R_{th} = 9 \Omega[/latex]

When the terminals A and B are open (i.e., load resistor [latex]R_{L}[/latex] is removed), the current flowing through the mesh ECDFE, as shown in Figure 5.106(b), is

[latex] I = \frac{20}{10 + 10} = 1 A[/latex]

Open-circuit voltage across AB, [latex]V_{AB} = V_{CD} = E_{th} = 10 × 1 = 10 V[/latex]
(since no current is flowing through 4-Ω resistor). Maximum power delivered to the load

[latex]P_{max} = I_{max}^{2} R_{L} = \left\lgroup\frac{E_{th}}{R_{th} + R_{L}} ight group^{2} imes R_{L} = \left\lgroup\frac{10}{9 + 9} ight group^{2} imes 9 = 2.778 W[/latex]

Question 5.47: In the network shown in Figure 5.105, determine the value of...

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