Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Question

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Answer 1

Question

Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Expert Solution & Answer

Answer to Problem 60P

The power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $io$ is the current flow through $1 Ω$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 60P

Apply Kirchhoff’s current law at node $v1$ in Figure 1.

$v1−01+v1−2804+0.5io=0$ (1)

With reference to Figure 1, write the expression for $io$ through $1 Ω$ branch.

$io=v1−01$ (2)

Substitute equation (2) in (1).

$v1−01+v1−2804+0.5(v1−01)=0v1+v14−2804+0.5v1=0v1+0.25v1+0.5v1=2804v1+0.25v1+0.5v1=70$

Simplify the equation as follows.

$v1+0.25v1+0.5v1=701.75v1=70v1=701.75v1=40 V$

Apply Kirchhoff’s current law at node $v2$ in Figure 1.

$v2−2808+v2−02−0.5io=0$ (3)

Substitute equation (2) in (3).

$v2−2808+v2−02−0.5(v1−01)=0v28−2808+v22−0.5v1=0−0.5v1+(18+12)v2=2808$

$−0.5v1+0.625v2=35$ (4)

Substitute $40 V$ for $v1$ in equation (4).

$−0.5(40)+0.625v2=35−20+0.625v2=350.625v2=35+200.625v2=55$

Simplify the equation as follows.

$v2=550.625=88 V$

Write the expression for power dissipated through $1 Ω$ resistor.

$P1 Ω=(v1−0)21 Ω$ (5)

Substitute $40 V$ for $v1$ in equation (5) to find $P1 Ω$ .

$P1 Ω=(40 V−0)21 Ω=1600 W$

Write the expression for power dissipated through $2 Ω$ resistor.

$P2 Ω=(v2−0)22 Ω$ (6)

Substitute $88 V$ for $v2$ in equation (6) to find $P2 Ω$ .

$P2 Ω=(88 V−0)22 Ω=7744 V22 Ω=3872 W$

Write the expression for power dissipated through $4 Ω$ resistor.

$P4 Ω=(v1−280)24 Ω$ (7)

Substitute $40 V$ for $v1$ in equation (7) to find $P4 Ω$ .

$P4 Ω=(40 V−280 V)24 Ω=(−240 V)24 Ω=57600 V24 Ω=14400 W$

Write the expression for power dissipated through $8 Ω$ resistor.

$P8 Ω=(v2−280 V)28 Ω$ (8)

Substitute $88 V$ for $v2$ in equation (8) to find $P8 Ω$ .

$P8 Ω=(88 V−280 V)28 Ω=(−192 V)28 Ω=36864 V28 Ω=4608 W$

Conclusion:

Therefore, the power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

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Answer 2

Question

Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Expert Solution & Answer

Answer to Problem 60P

The power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $io$ is the current flow through $1 Ω$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 60P

Apply Kirchhoff’s current law at node $v1$ in Figure 1.

$v1−01+v1−2804+0.5io=0$ (1)

With reference to Figure 1, write the expression for $io$ through $1 Ω$ branch.

$io=v1−01$ (2)

Substitute equation (2) in (1).

$v1−01+v1−2804+0.5(v1−01)=0v1+v14−2804+0.5v1=0v1+0.25v1+0.5v1=2804v1+0.25v1+0.5v1=70$

Simplify the equation as follows.

$v1+0.25v1+0.5v1=701.75v1=70v1=701.75v1=40 V$

Apply Kirchhoff’s current law at node $v2$ in Figure 1.

$v2−2808+v2−02−0.5io=0$ (3)

Substitute equation (2) in (3).

$v2−2808+v2−02−0.5(v1−01)=0v28−2808+v22−0.5v1=0−0.5v1+(18+12)v2=2808$

$−0.5v1+0.625v2=35$ (4)

Substitute $40 V$ for $v1$ in equation (4).

$−0.5(40)+0.625v2=35−20+0.625v2=350.625v2=35+200.625v2=55$

Simplify the equation as follows.

$v2=550.625=88 V$

Write the expression for power dissipated through $1 Ω$ resistor.

$P1 Ω=(v1−0)21 Ω$ (5)

Substitute $40 V$ for $v1$ in equation (5) to find $P1 Ω$ .

$P1 Ω=(40 V−0)21 Ω=1600 W$

Write the expression for power dissipated through $2 Ω$ resistor.

$P2 Ω=(v2−0)22 Ω$ (6)

Substitute $88 V$ for $v2$ in equation (6) to find $P2 Ω$ .

$P2 Ω=(88 V−0)22 Ω=7744 V22 Ω=3872 W$

Write the expression for power dissipated through $4 Ω$ resistor.

$P4 Ω=(v1−280)24 Ω$ (7)

Substitute $40 V$ for $v1$ in equation (7) to find $P4 Ω$ .

$P4 Ω=(40 V−280 V)24 Ω=(−240 V)24 Ω=57600 V24 Ω=14400 W$

Write the expression for power dissipated through $8 Ω$ resistor.

$P8 Ω=(v2−280 V)28 Ω$ (8)

Substitute $88 V$ for $v2$ in equation (8) to find $P8 Ω$ .

$P8 Ω=(88 V−280 V)28 Ω=(−192 V)28 Ω=36864 V28 Ω=4608 W$

Conclusion:

Therefore, the power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

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Answer 3

Question

Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Expert Solution & Answer

Answer to Problem 60P

The power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $io$ is the current flow through $1 Ω$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 60P

Apply Kirchhoff’s current law at node $v1$ in Figure 1.

$v1−01+v1−2804+0.5io=0$ (1)

With reference to Figure 1, write the expression for $io$ through $1 Ω$ branch.

$io=v1−01$ (2)

Substitute equation (2) in (1).

$v1−01+v1−2804+0.5(v1−01)=0v1+v14−2804+0.5v1=0v1+0.25v1+0.5v1=2804v1+0.25v1+0.5v1=70$

Simplify the equation as follows.

$v1+0.25v1+0.5v1=701.75v1=70v1=701.75v1=40 V$

Apply Kirchhoff’s current law at node $v2$ in Figure 1.

$v2−2808+v2−02−0.5io=0$ (3)

Substitute equation (2) in (3).

$v2−2808+v2−02−0.5(v1−01)=0v28−2808+v22−0.5v1=0−0.5v1+(18+12)v2=2808$

$−0.5v1+0.625v2=35$ (4)

Substitute $40 V$ for $v1$ in equation (4).

$−0.5(40)+0.625v2=35−20+0.625v2=350.625v2=35+200.625v2=55$

Simplify the equation as follows.

$v2=550.625=88 V$

Write the expression for power dissipated through $1 Ω$ resistor.

$P1 Ω=(v1−0)21 Ω$ (5)

Substitute $40 V$ for $v1$ in equation (5) to find $P1 Ω$ .

$P1 Ω=(40 V−0)21 Ω=1600 W$

Write the expression for power dissipated through $2 Ω$ resistor.

$P2 Ω=(v2−0)22 Ω$ (6)

Substitute $88 V$ for $v2$ in equation (6) to find $P2 Ω$ .

$P2 Ω=(88 V−0)22 Ω=7744 V22 Ω=3872 W$

Write the expression for power dissipated through $4 Ω$ resistor.

$P4 Ω=(v1−280)24 Ω$ (7)

Substitute $40 V$ for $v1$ in equation (7) to find $P4 Ω$ .

$P4 Ω=(40 V−280 V)24 Ω=(−240 V)24 Ω=57600 V24 Ω=14400 W$

Write the expression for power dissipated through $8 Ω$ resistor.

$P8 Ω=(v2−280 V)28 Ω$ (8)

Substitute $88 V$ for $v2$ in equation (8) to find $P8 Ω$ .

$P8 Ω=(88 V−280 V)28 Ω=(−192 V)28 Ω=36864 V28 Ω=4608 W$

Conclusion:

Therefore, the power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

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Answer 4

Question

Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Expert Solution & Answer

Answer to Problem 60P

The power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $io$ is the current flow through $1 Ω$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 60P

Apply Kirchhoff’s current law at node $v1$ in Figure 1.

$v1−01+v1−2804+0.5io=0$ (1)

With reference to Figure 1, write the expression for $io$ through $1 Ω$ branch.

$io=v1−01$ (2)

Substitute equation (2) in (1).

$v1−01+v1−2804+0.5(v1−01)=0v1+v14−2804+0.5v1=0v1+0.25v1+0.5v1=2804v1+0.25v1+0.5v1=70$

Simplify the equation as follows.

$v1+0.25v1+0.5v1=701.75v1=70v1=701.75v1=40 V$

Apply Kirchhoff’s current law at node $v2$ in Figure 1.

$v2−2808+v2−02−0.5io=0$ (3)

Substitute equation (2) in (3).

$v2−2808+v2−02−0.5(v1−01)=0v28−2808+v22−0.5v1=0−0.5v1+(18+12)v2=2808$

$−0.5v1+0.625v2=35$ (4)

Substitute $40 V$ for $v1$ in equation (4).

$−0.5(40)+0.625v2=35−20+0.625v2=350.625v2=35+200.625v2=55$

Simplify the equation as follows.

$v2=550.625=88 V$

Write the expression for power dissipated through $1 Ω$ resistor.

$P1 Ω=(v1−0)21 Ω$ (5)

Substitute $40 V$ for $v1$ in equation (5) to find $P1 Ω$ .

$P1 Ω=(40 V−0)21 Ω=1600 W$

Write the expression for power dissipated through $2 Ω$ resistor.

$P2 Ω=(v2−0)22 Ω$ (6)

Substitute $88 V$ for $v2$ in equation (6) to find $P2 Ω$ .

$P2 Ω=(88 V−0)22 Ω=7744 V22 Ω=3872 W$

Write the expression for power dissipated through $4 Ω$ resistor.

$P4 Ω=(v1−280)24 Ω$ (7)

Substitute $40 V$ for $v1$ in equation (7) to find $P4 Ω$ .

$P4 Ω=(40 V−280 V)24 Ω=(−240 V)24 Ω=57600 V24 Ω=14400 W$

Write the expression for power dissipated through $8 Ω$ resistor.

$P8 Ω=(v2−280 V)28 Ω$ (8)

Substitute $88 V$ for $v2$ in equation (8) to find $P8 Ω$ .

$P8 Ω=(88 V−280 V)28 Ω=(−192 V)28 Ω=36864 V28 Ω=4608 W$

Conclusion:

Therefore, the power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

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Answer 5

Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Expert Solution & Answer

Answer to Problem 60P

The power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $io$ is the current flow through $1 Ω$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 60P

Apply Kirchhoff’s current law at node $v1$ in Figure 1.

$v1−01+v1−2804+0.5io=0$ (1)

With reference to Figure 1, write the expression for $io$ through $1 Ω$ branch.

$io=v1−01$ (2)

Substitute equation (2) in (1).

$v1−01+v1−2804+0.5(v1−01)=0v1+v14−2804+0.5v1=0v1+0.25v1+0.5v1=2804v1+0.25v1+0.5v1=70$

Simplify the equation as follows.

$v1+0.25v1+0.5v1=701.75v1=70v1=701.75v1=40 V$

Apply Kirchhoff’s current law at node $v2$ in Figure 1.

$v2−2808+v2−02−0.5io=0$ (3)

Substitute equation (2) in (3).

$v2−2808+v2−02−0.5(v1−01)=0v28−2808+v22−0.5v1=0−0.5v1+(18+12)v2=2808$

$−0.5v1+0.625v2=35$ (4)

Substitute $40 V$ for $v1$ in equation (4).

$−0.5(40)+0.625v2=35−20+0.625v2=350.625v2=35+200.625v2=55$

Simplify the equation as follows.

$v2=550.625=88 V$

Write the expression for power dissipated through $1 Ω$ resistor.

$P1 Ω=(v1−0)21 Ω$ (5)

Substitute $40 V$ for $v1$ in equation (5) to find $P1 Ω$ .

$P1 Ω=(40 V−0)21 Ω=1600 W$

Write the expression for power dissipated through $2 Ω$ resistor.

$P2 Ω=(v2−0)22 Ω$ (6)

Substitute $88 V$ for $v2$ in equation (6) to find $P2 Ω$ .

$P2 Ω=(88 V−0)22 Ω=7744 V22 Ω=3872 W$

Write the expression for power dissipated through $4 Ω$ resistor.

$P4 Ω=(v1−280)24 Ω$ (7)

Substitute $40 V$ for $v1$ in equation (7) to find $P4 Ω$ .

$P4 Ω=(40 V−280 V)24 Ω=(−240 V)24 Ω=57600 V24 Ω=14400 W$

Write the expression for power dissipated through $8 Ω$ resistor.

$P8 Ω=(v2−280 V)28 Ω$ (8)

Substitute $88 V$ for $v2$ in equation (8) to find $P8 Ω$ .

$P8 Ω=(88 V−280 V)28 Ω=(−192 V)28 Ω=36864 V28 Ω=4608 W$

Conclusion:

Therefore, the power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Answer 6

Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Expert Solution & Answer

Answer to Problem 60P

The power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $io$ is the current flow through $1 Ω$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 60P

Apply Kirchhoff’s current law at node $v1$ in Figure 1.

$v1−01+v1−2804+0.5io=0$ (1)

With reference to Figure 1, write the expression for $io$ through $1 Ω$ branch.

$io=v1−01$ (2)

Substitute equation (2) in (1).

$v1−01+v1−2804+0.5(v1−01)=0v1+v14−2804+0.5v1=0v1+0.25v1+0.5v1=2804v1+0.25v1+0.5v1=70$

Simplify the equation as follows.

$v1+0.25v1+0.5v1=701.75v1=70v1=701.75v1=40 V$

Apply Kirchhoff’s current law at node $v2$ in Figure 1.

$v2−2808+v2−02−0.5io=0$ (3)

Substitute equation (2) in (3).

$v2−2808+v2−02−0.5(v1−01)=0v28−2808+v22−0.5v1=0−0.5v1+(18+12)v2=2808$

$−0.5v1+0.625v2=35$ (4)

Substitute $40 V$ for $v1$ in equation (4).

$−0.5(40)+0.625v2=35−20+0.625v2=350.625v2=35+200.625v2=55$

Simplify the equation as follows.

$v2=550.625=88 V$

Write the expression for power dissipated through $1 Ω$ resistor.

$P1 Ω=(v1−0)21 Ω$ (5)

Substitute $40 V$ for $v1$ in equation (5) to find $P1 Ω$ .

$P1 Ω=(40 V−0)21 Ω=1600 W$

Write the expression for power dissipated through $2 Ω$ resistor.

$P2 Ω=(v2−0)22 Ω$ (6)

Substitute $88 V$ for $v2$ in equation (6) to find $P2 Ω$ .

$P2 Ω=(88 V−0)22 Ω=7744 V22 Ω=3872 W$

Write the expression for power dissipated through $4 Ω$ resistor.

$P4 Ω=(v1−280)24 Ω$ (7)

Substitute $40 V$ for $v1$ in equation (7) to find $P4 Ω$ .

$P4 Ω=(40 V−280 V)24 Ω=(−240 V)24 Ω=57600 V24 Ω=14400 W$

Write the expression for power dissipated through $8 Ω$ resistor.

$P8 Ω=(v2−280 V)28 Ω$ (8)

Substitute $88 V$ for $v2$ in equation (8) to find $P8 Ω$ .

$P8 Ω=(88 V−280 V)28 Ω=(−192 V)28 Ω=36864 V28 Ω=4608 W$

Conclusion:

Therefore, the power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Answer 7

Chapter 3, Problem 60P

To determine

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Answer 8

Expert Solution & Answer

Answer to Problem 60P

The power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

Refer to Figure 3.104 in the textbook for the nodal analysis. In the given circuit, $io$ is the current flow through $1 Ω$ resistor.

Calculation:

Modify the given figure with the representation of node voltage as shown in Figure 1.

EBK FUNDAMENTALS OF ELECTRIC CIRCUITS, Chapter 3, Problem 60P

Apply Kirchhoff’s current law at node $v1$ in Figure 1.

$v1−01+v1−2804+0.5io=0$ (1)

With reference to Figure 1, write the expression for $io$ through $1 Ω$ branch.

$io=v1−01$ (2)

Substitute equation (2) in (1).

$v1−01+v1−2804+0.5(v1−01)=0v1+v14−2804+0.5v1=0v1+0.25v1+0.5v1=2804v1+0.25v1+0.5v1=70$

Simplify the equation as follows.

$v1+0.25v1+0.5v1=701.75v1=70v1=701.75v1=40 V$

Apply Kirchhoff’s current law at node $v2$ in Figure 1.

$v2−2808+v2−02−0.5io=0$ (3)

Substitute equation (2) in (3).

$v2−2808+v2−02−0.5(v1−01)=0v28−2808+v22−0.5v1=0−0.5v1+(18+12)v2=2808$

$−0.5v1+0.625v2=35$ (4)

Substitute $40 V$ for $v1$ in equation (4).

$−0.5(40)+0.625v2=35−20+0.625v2=350.625v2=35+200.625v2=55$

Simplify the equation as follows.

$v2=550.625=88 V$

Write the expression for power dissipated through $1 Ω$ resistor.

$P1 Ω=(v1−0)21 Ω$ (5)

Substitute $40 V$ for $v1$ in equation (5) to find $P1 Ω$ .

$P1 Ω=(40 V−0)21 Ω=1600 W$

Write the expression for power dissipated through $2 Ω$ resistor.

$P2 Ω=(v2−0)22 Ω$ (6)

Substitute $88 V$ for $v2$ in equation (6) to find $P2 Ω$ .

$P2 Ω=(88 V−0)22 Ω=7744 V22 Ω=3872 W$

Write the expression for power dissipated through $4 Ω$ resistor.

$P4 Ω=(v1−280)24 Ω$ (7)

Substitute $40 V$ for $v1$ in equation (7) to find $P4 Ω$ .

$P4 Ω=(40 V−280 V)24 Ω=(−240 V)24 Ω=57600 V24 Ω=14400 W$

Write the expression for power dissipated through $8 Ω$ resistor.

$P8 Ω=(v2−280 V)28 Ω$ (8)

Substitute $88 V$ for $v2$ in equation (8) to find $P8 Ω$ .

$P8 Ω=(88 V−280 V)28 Ω=(−192 V)28 Ω=36864 V28 Ω=4608 W$

Conclusion:

Therefore, the power dissipated in $1 Ω$ , $2 Ω$ , $4 Ω$ , and $8 Ω$ resistors are $1600 W$ , $3872 W$ , $14400 W$ , and $4608 W$ respectively.

Answer 12

Ch. 3.2 - Figure 3.4 For Practice Prob. 3.1. Obtain the node...Ch. 3.2 - Figure 3.6 For Practice Prob. 3.2. Find the...Ch. 3.3 - Figure 3.11 For Practice Prob. 3.3. Find v and i...Ch. 3.3 - Figure 3.14 For Practice Prob. 3.4. Find v1, v2,...Ch. 3.4 - Practice Problem 3.5 Figure 3.19 For Practice...Ch. 3.4 - Practice Problem 3.6 Figure 3.21 For Practice...Ch. 3.5 - Practice Problem 3.7 Figure 3.25 For Practice...Ch. 3.6 - By inspection, obtain the node-voltage equations...Ch. 3.6 - By inspection, obtain the mesh-current equations...Ch. 3.8 - For the circuit in Fig. 3.33, use PSpice to find...

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Concept explainers

Calculate the power dissipated in each resistor in the circuit of Figure 3.104.

Answer to Problem 60P

Explanation of Solution

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Chapter 3 Solutions