Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
6th Edition
ISBN: 9780078028229
Author: Charles K Alexander, Matthew Sadiku
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 4, Problem 69P

Find the maximum power transferred to resistor R in the circuit of Fig. 4.135.

Chapter 4, Problem 69P, Find the maximum power transferred to resistor R in the circuit of Fig. 4.135. Figure 4.135

Figure 4.135

Expert Solution & Answer
Check Mark
To determine

Calculate the maximum power delivered to the resistor R of the circuit shown in Figure 4.135.

Answer to Problem 69P

The maximum power delivered to the variable resistor R is infinity.

Explanation of Solution

Given data:

Refer to Figure 4.135 in the textbook.

The voltage source is 100 V.

Formula used:

Write the expression to find the power delivered to the resistor.

pmax=VTh24RTh        (1)

Here,

VTh is the Thevenin voltage, and

RTh is the Thevenin resistance.

Calculation:

The given circuit is modified as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 4, Problem 69P , additional homework tip  1

In Figure 1, the voltage source with series resistance is converted into current source with parallel resistance using source transformation. The current (I) is calculated by using Ohm’s law.

I=100V10×103Ω{1k=103}=10×103A{1A=1V1Ω}=10mA   {1m=103} 

The source transformation is shown in Figure 2.

Fundamentals of Electric Circuits, Chapter 4, Problem 69P , additional homework tip  2

In Figure 2, apply Kirchhoff’s current law at node voltage v1 as follows.

(v1v222×103)(0.006v2)+(v1030×103)=0{vo=v2,1k=103}(0.0454545×103)(v1v2)(0.006v2)+(0.0333×103)v1=0(0.0454545×103)v1(0.0454545×103)v2(0.006v2)+(0.03333×103)v1=0(7.87845×105)v1(6.0454545×103)v2=0

Simplify the equation as follows,

v1=(6.0454545×103)v2(7.87845×105)

v1=76.734v2        (2)

In Figure 2, apply Kirchhoff’s current law at node voltage v2 as follows.

0.01+(v2010×103)+(v2040×103)+(v2v122×103)=0{1k=103}         (3)

0.01+(0.1×103)v2+(0.025×103)v2+(0.0454545×103)v2(0.0454545×103)v1=0(1.704545×104)v2(0.0454545×103)v1=0.01

Substitute 76.734v2 for v1 to find the node voltage v2 in volts.

(1.704545×104)v2(0.0454545×103)(76.734v2)=0.01(1.704545×104)v2(3.4879×103)v2=0.01(3.317445×103)v2=0.01

Simplify the equation as follows,

v2=3.0143V

Substitute 3.0143V for v2 in equation (2) to find the node voltage v1 in volts.

v1=76.734(3.0143V)v1=231.3V

Since, the voltage at node v1 is the open circuit voltage. Therefore,

voc=v1=231.3V

Refer to Figure 4.135 in the textbook.

In the given circuit, find the short circuit current by shorting the resistor R.

The modified circuit is shown in Figure 3.

Fundamentals of Electric Circuits, Chapter 4, Problem 69P , additional homework tip  3

To find the short circuit current isc, set v1=0V.

Substitute 0 for v1 in equation (3) to find the node voltage v2 in volts.

0.01+(v2010×103)+(v2040×103)+(v2022×103)=00.01+(0.1×103)v2+(0.025×103)v2+(0.0454545×103)v2=0(1.704545×104)v2=0.01v2=58.6667V

In Figure 3, apply Kirchhoff’s current law at node voltage v1 as follows.

(v1v222×103)0.006v2+isc=0{vo=v2}isc=0.006v2(v1v222×103)

Substitute 0 for v1 and 58.6667 for v2 to find the short circuit current isc in amperes.

isc=0.006(58.6667)(058.666722×103)=0.352+(2.666×103)=0.35467A

The Thevenin resistance is,

RTh=vocisc

Substitute 231.3 for voc and 0.35467 for isc to find the Thevenin resistance in ohms.

RTh=231.30.35467=652.2Ω

The negative equivalent resistance indicates that an active device (dependent source) presents in the circuit, since the circuit cannot have a negative resistance in a purely passive circuit. The negative resistance for the equivalent circuit means that both the resistance and the source will effectively delivers the power to load.

Since the resistance cannot be the negative, the correct answer will be R=652.2Ω which then means the current through the resistor R is infinite and the power delivered to the resistor R is infinity.

Conclusion:

Thus, the maximum power delivered to the variable resistor R is infinity.

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