The current in each resistor

Question

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Answer 1

Question

Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

(a)

Expert Solution

Answer to Problem 90P

The current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Explanation of Solution

Given:

The voltage of the battery is ε=8 V .

The circuit is shown in figure.

Physics for Scientists and Engineers, Chapter 25, Problem 90P

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

Iin=Iout

The expression for the Kirchhoff’s loop rule is given by:

∑V=0

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(6 Ω×I 6 Ω)−(2 Ω×I 1,2 Ω)=03 Ω×I1,2 Ω+6 Ω×I6 Ω=12 V ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(2 Ω×I 2 Ω)−(2 Ω×I 1,2 Ω)−4 V=03 Ω×I1,2 Ω+2 Ω×I2 Ω=8 V

Apply the Kirchhoff’s junction rule,

I6 Ω=I2 Ω+I1,2 Ω3 Ω×I1,2 Ω+6 Ω×(I 2 Ω+I 1,2 Ω)=12 V (2)

On solving (1) and (2) equation,

I1,2 Ω=2 AI2 Ω=−1 AI6 Ω=2 A+(−1 A)=1 A

Conclusion:

Therefore, the current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

(b)

To determine

The power supplied by each source of emf.

(b)

Expert Solution

Answer to Problem 90P

The power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

P=IV

Calculation:

The power delivered by the source of emf 8 V is calculated as,

P=IVP8 V=I1,2 Ω×8 V=(2 A)×(8 V)=16 W

The power delivered by source of 4V is calculated as,

P4 V=I2 Ω×4 V=(−1 A)×(4 V)=−4 W

Conclusion:

Therefore, the power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

(c)

To determine

The power delivered to each resistor

(c)

Expert Solution

Answer to Problem 90P

The power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

P=IR2

Calculation:

The delivered by the 1 Ω resistor is given by

P1 Ω=I1,2 Ω( R 1 Ω)2=(2 A)(1 Ω)=2 W

Similarly,

P2 Ω=I1,2 Ω( R 2 Ω)2=(2 A)(2 Ω)=4 W

P2 Ω=I2 Ω( R 2 Ω)2=(−1 A)(2 Ω)=−2 W

P6 Ω=I6 Ω( R 6 Ω)2=(1 A)(6 Ω)=6 W

Conclusion:

Therefore, the power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

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Answer 2

Question

Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

(a)

Expert Solution

Answer to Problem 90P

The current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Explanation of Solution

Given:

The voltage of the battery is ε=8 V .

The circuit is shown in figure.

Physics for Scientists and Engineers, Chapter 25, Problem 90P

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

Iin=Iout

The expression for the Kirchhoff’s loop rule is given by:

∑V=0

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(6 Ω×I 6 Ω)−(2 Ω×I 1,2 Ω)=03 Ω×I1,2 Ω+6 Ω×I6 Ω=12 V ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(2 Ω×I 2 Ω)−(2 Ω×I 1,2 Ω)−4 V=03 Ω×I1,2 Ω+2 Ω×I2 Ω=8 V

Apply the Kirchhoff’s junction rule,

I6 Ω=I2 Ω+I1,2 Ω3 Ω×I1,2 Ω+6 Ω×(I 2 Ω+I 1,2 Ω)=12 V (2)

On solving (1) and (2) equation,

I1,2 Ω=2 AI2 Ω=−1 AI6 Ω=2 A+(−1 A)=1 A

Conclusion:

Therefore, the current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

(b)

To determine

The power supplied by each source of emf.

(b)

Expert Solution

Answer to Problem 90P

The power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

P=IV

Calculation:

The power delivered by the source of emf 8 V is calculated as,

P=IVP8 V=I1,2 Ω×8 V=(2 A)×(8 V)=16 W

The power delivered by source of 4V is calculated as,

P4 V=I2 Ω×4 V=(−1 A)×(4 V)=−4 W

Conclusion:

Therefore, the power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

(c)

To determine

The power delivered to each resistor

(c)

Expert Solution

Answer to Problem 90P

The power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

P=IR2

Calculation:

The delivered by the 1 Ω resistor is given by

P1 Ω=I1,2 Ω( R 1 Ω)2=(2 A)(1 Ω)=2 W

Similarly,

P2 Ω=I1,2 Ω( R 2 Ω)2=(2 A)(2 Ω)=4 W

P2 Ω=I2 Ω( R 2 Ω)2=(−1 A)(2 Ω)=−2 W

P6 Ω=I6 Ω( R 6 Ω)2=(1 A)(6 Ω)=6 W

Conclusion:

Therefore, the power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

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Answer 3

Question

Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

(a)

Expert Solution

Answer to Problem 90P

The current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Explanation of Solution

Given:

The voltage of the battery is ε=8 V .

The circuit is shown in figure.

Physics for Scientists and Engineers, Chapter 25, Problem 90P

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

Iin=Iout

The expression for the Kirchhoff’s loop rule is given by:

∑V=0

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(6 Ω×I 6 Ω)−(2 Ω×I 1,2 Ω)=03 Ω×I1,2 Ω+6 Ω×I6 Ω=12 V ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(2 Ω×I 2 Ω)−(2 Ω×I 1,2 Ω)−4 V=03 Ω×I1,2 Ω+2 Ω×I2 Ω=8 V

Apply the Kirchhoff’s junction rule,

I6 Ω=I2 Ω+I1,2 Ω3 Ω×I1,2 Ω+6 Ω×(I 2 Ω+I 1,2 Ω)=12 V (2)

On solving (1) and (2) equation,

I1,2 Ω=2 AI2 Ω=−1 AI6 Ω=2 A+(−1 A)=1 A

Conclusion:

Therefore, the current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

(b)

To determine

The power supplied by each source of emf.

(b)

Expert Solution

Answer to Problem 90P

The power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

P=IV

Calculation:

The power delivered by the source of emf 8 V is calculated as,

P=IVP8 V=I1,2 Ω×8 V=(2 A)×(8 V)=16 W

The power delivered by source of 4V is calculated as,

P4 V=I2 Ω×4 V=(−1 A)×(4 V)=−4 W

Conclusion:

Therefore, the power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

(c)

To determine

The power delivered to each resistor

(c)

Expert Solution

Answer to Problem 90P

The power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

P=IR2

Calculation:

The delivered by the 1 Ω resistor is given by

P1 Ω=I1,2 Ω( R 1 Ω)2=(2 A)(1 Ω)=2 W

Similarly,

P2 Ω=I1,2 Ω( R 2 Ω)2=(2 A)(2 Ω)=4 W

P2 Ω=I2 Ω( R 2 Ω)2=(−1 A)(2 Ω)=−2 W

P6 Ω=I6 Ω( R 6 Ω)2=(1 A)(6 Ω)=6 W

Conclusion:

Therefore, the power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

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Answer 4

Question

Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

(a)

Expert Solution

Answer to Problem 90P

The current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Explanation of Solution

Given:

The voltage of the battery is ε=8 V .

The circuit is shown in figure.

Physics for Scientists and Engineers, Chapter 25, Problem 90P

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

Iin=Iout

The expression for the Kirchhoff’s loop rule is given by:

∑V=0

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(6 Ω×I 6 Ω)−(2 Ω×I 1,2 Ω)=03 Ω×I1,2 Ω+6 Ω×I6 Ω=12 V ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(2 Ω×I 2 Ω)−(2 Ω×I 1,2 Ω)−4 V=03 Ω×I1,2 Ω+2 Ω×I2 Ω=8 V

Apply the Kirchhoff’s junction rule,

I6 Ω=I2 Ω+I1,2 Ω3 Ω×I1,2 Ω+6 Ω×(I 2 Ω+I 1,2 Ω)=12 V (2)

On solving (1) and (2) equation,

I1,2 Ω=2 AI2 Ω=−1 AI6 Ω=2 A+(−1 A)=1 A

Conclusion:

Therefore, the current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

(b)

To determine

The power supplied by each source of emf.

(b)

Expert Solution

Answer to Problem 90P

The power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

P=IV

Calculation:

The power delivered by the source of emf 8 V is calculated as,

P=IVP8 V=I1,2 Ω×8 V=(2 A)×(8 V)=16 W

The power delivered by source of 4V is calculated as,

P4 V=I2 Ω×4 V=(−1 A)×(4 V)=−4 W

Conclusion:

Therefore, the power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

(c)

To determine

The power delivered to each resistor

(c)

Expert Solution

Answer to Problem 90P

The power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

P=IR2

Calculation:

The delivered by the 1 Ω resistor is given by

P1 Ω=I1,2 Ω( R 1 Ω)2=(2 A)(1 Ω)=2 W

Similarly,

P2 Ω=I1,2 Ω( R 2 Ω)2=(2 A)(2 Ω)=4 W

P2 Ω=I2 Ω( R 2 Ω)2=(−1 A)(2 Ω)=−2 W

P6 Ω=I6 Ω( R 6 Ω)2=(1 A)(6 Ω)=6 W

Conclusion:

Therefore, the power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

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Answer 5

Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

(a)

Expert Solution

Answer to Problem 90P

The current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Explanation of Solution

Given:

The voltage of the battery is ε=8 V .

The circuit is shown in figure.

Physics for Scientists and Engineers, Chapter 25, Problem 90P

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

Iin=Iout

The expression for the Kirchhoff’s loop rule is given by:

∑V=0

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(6 Ω×I 6 Ω)−(2 Ω×I 1,2 Ω)=03 Ω×I1,2 Ω+6 Ω×I6 Ω=12 V ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(2 Ω×I 2 Ω)−(2 Ω×I 1,2 Ω)−4 V=03 Ω×I1,2 Ω+2 Ω×I2 Ω=8 V

Apply the Kirchhoff’s junction rule,

I6 Ω=I2 Ω+I1,2 Ω3 Ω×I1,2 Ω+6 Ω×(I 2 Ω+I 1,2 Ω)=12 V (2)

On solving (1) and (2) equation,

I1,2 Ω=2 AI2 Ω=−1 AI6 Ω=2 A+(−1 A)=1 A

Conclusion:

Therefore, the current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

(b)

To determine

The power supplied by each source of emf.

(b)

Expert Solution

Answer to Problem 90P

The power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

P=IV

Calculation:

The power delivered by the source of emf 8 V is calculated as,

P=IVP8 V=I1,2 Ω×8 V=(2 A)×(8 V)=16 W

The power delivered by source of 4V is calculated as,

P4 V=I2 Ω×4 V=(−1 A)×(4 V)=−4 W

Conclusion:

Therefore, the power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

(c)

To determine

The power delivered to each resistor

(c)

Expert Solution

Answer to Problem 90P

The power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

P=IR2

Calculation:

The delivered by the 1 Ω resistor is given by

P1 Ω=I1,2 Ω( R 1 Ω)2=(2 A)(1 Ω)=2 W

Similarly,

P2 Ω=I1,2 Ω( R 2 Ω)2=(2 A)(2 Ω)=4 W

P2 Ω=I2 Ω( R 2 Ω)2=(−1 A)(2 Ω)=−2 W

P6 Ω=I6 Ω( R 6 Ω)2=(1 A)(6 Ω)=6 W

Conclusion:

Therefore, the power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Answer 6

Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

(a)

Expert Solution

Answer to Problem 90P

The current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Explanation of Solution

Given:

The voltage of the battery is ε=8 V .

The circuit is shown in figure.

Physics for Scientists and Engineers, Chapter 25, Problem 90P

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

Iin=Iout

The expression for the Kirchhoff’s loop rule is given by:

∑V=0

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(6 Ω×I 6 Ω)−(2 Ω×I 1,2 Ω)=03 Ω×I1,2 Ω+6 Ω×I6 Ω=12 V ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(2 Ω×I 2 Ω)−(2 Ω×I 1,2 Ω)−4 V=03 Ω×I1,2 Ω+2 Ω×I2 Ω=8 V

Apply the Kirchhoff’s junction rule,

I6 Ω=I2 Ω+I1,2 Ω3 Ω×I1,2 Ω+6 Ω×(I 2 Ω+I 1,2 Ω)=12 V (2)

On solving (1) and (2) equation,

I1,2 Ω=2 AI2 Ω=−1 AI6 Ω=2 A+(−1 A)=1 A

Conclusion:

Therefore, the current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Answer 7

Chapter 25, Problem 90P

(a)

To determine

The current in each resistor

Answer 8

(a)

Expert Solution

Answer to Problem 90P

The current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The voltage of the battery is ε=8 V .

The circuit is shown in figure.

Physics for Scientists and Engineers, Chapter 25, Problem 90P

Figure (1)

Formula used:

The expression for the Kirchhoff’s junction ruleis given by,

Iin=Iout

The expression for the Kirchhoff’s loop rule is given by:

∑V=0

Calculation:

Applying the Kirchhoff’s loop rule to the outside loop of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(6 Ω×I 6 Ω)−(2 Ω×I 1,2 Ω)=03 Ω×I1,2 Ω+6 Ω×I6 Ω=12 V ..... (1)

Similarly, apply the Kirchhoff’s loop rule to the inside loop at the LHS of the circuit:

8 V+4 V−(1 Ω×I 1,2 Ω)−(2 Ω×I 2 Ω)−(2 Ω×I 1,2 Ω)−4 V=03 Ω×I1,2 Ω+2 Ω×I2 Ω=8 V

Apply the Kirchhoff’s junction rule,

I6 Ω=I2 Ω+I1,2 Ω3 Ω×I1,2 Ω+6 Ω×(I 2 Ω+I 1,2 Ω)=12 V (2)

On solving (1) and (2) equation,

I1,2 Ω=2 AI2 Ω=−1 AI6 Ω=2 A+(−1 A)=1 A

Conclusion:

Therefore, the current in each resistor is I1,2 Ω=2 A, I2 Ω=−1 A and I6 Ω=1 A .

Answer 13

(b)

To determine

The power supplied by each source of emf.

(b)

Expert Solution

Answer to Problem 90P

The power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

P=IV

Calculation:

The power delivered by the source of emf 8 V is calculated as,

P=IVP8 V=I1,2 Ω×8 V=(2 A)×(8 V)=16 W

The power delivered by source of 4V is calculated as,

P4 V=I2 Ω×4 V=(−1 A)×(4 V)=−4 W

Conclusion:

Therefore, the power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Answer 14

(b)

To determine

The power supplied by each source of emf.

Answer 15

(b)

Expert Solution

Answer to Problem 90P

The power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The expression for the power supplied by emf source is given by,

P=IV

Calculation:

The power delivered by the source of emf 8 V is calculated as,

P=IVP8 V=I1,2 Ω×8 V=(2 A)×(8 V)=16 W

The power delivered by source of 4V is calculated as,

P4 V=I2 Ω×4 V=(−1 A)×(4 V)=−4 W

Conclusion:

Therefore, the power supplied by each source of emf is P8 V=16 W and P4 V=−4 W .

Answer 20

(c)

To determine

The power delivered to each resistor

(c)

Expert Solution

Answer to Problem 90P

The power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

P=IR2

Calculation:

The delivered by the 1 Ω resistor is given by

P1 Ω=I1,2 Ω( R 1 Ω)2=(2 A)(1 Ω)=2 W

Similarly,

P2 Ω=I1,2 Ω( R 2 Ω)2=(2 A)(2 Ω)=4 W

P2 Ω=I2 Ω( R 2 Ω)2=(−1 A)(2 Ω)=−2 W

P6 Ω=I6 Ω( R 6 Ω)2=(1 A)(6 Ω)=6 W

Conclusion:

Therefore, the power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Answer 21

(c)

To determine

The power delivered to each resistor

Answer 22

(c)

Expert Solution

Answer to Problem 90P

The power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula used:

The expression for the power delivered by the resistoris given by,

P=IR2

Calculation:

The delivered by the 1 Ω resistor is given by

P1 Ω=I1,2 Ω( R 1 Ω)2=(2 A)(1 Ω)=2 W

Similarly,

P2 Ω=I1,2 Ω( R 2 Ω)2=(2 A)(2 Ω)=4 W

P2 Ω=I2 Ω( R 2 Ω)2=(−1 A)(2 Ω)=−2 W

P6 Ω=I6 Ω( R 6 Ω)2=(1 A)(6 Ω)=6 W

Conclusion:

Therefore, the power delivered to each resistor is P1 Ω=2 W , P2 Ω=4 W , P2 Ω=−2 W and P6 Ω=6 W .

Answer 27

Ch. 25 - Prob. 1P Ch. 25 - Prob. 2P Ch. 25 - Prob. 3P Ch. 25 - Prob. 4P Ch. 25 - Prob. 5P Ch. 25 - Prob. 6P Ch. 25 - Prob. 7P Ch. 25 - Prob. 8P Ch. 25 - Prob. 9P Ch. 25 - Prob. 10P

The current in each resistor

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The current in each resistor

Answer to Problem 90P

Explanation of Solution

The power supplied by each source of emf.

Answer to Problem 90P

Explanation of Solution

The power delivered to each resistor

Answer to Problem 90P

Explanation of Solution

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