The value of current in battery.

Question

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Answer 1

Question

Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

(a)

Expert Solution

Answer to Problem 100P

The value of current in battery is $25.00 A$ .

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00 mC$ .

The value of capacitance is $C=5.00 μF$ .

The emf is $ε=310 V$ .

The current in $50.0 Ω$ resistor is $I50.0 Ω=5.00 A$ .

The current in $R2$ resistor is $IR2=5.00 A$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

$Ibattery=I10.0 Ω+I50.0 Ω$ ..... (1)

Calculation:

The expression for current in $10.0 Ω$ resistor is,

$I10.0 Ω=V 10.0 Ω10.0 Ω=1( 10.0 Ω)(qC) (∵V 10 Ω=qC)$ ..... (2)

From equation (1) and (2),

$Ibattery=1(10.0 Ω)(qC)+I50.0 Ω$

The current in battery is calculated as,

$Ibattery=1( 10.0 Ω)( 1.00 mC 5.00 μF)+(5.00 A)=1( 10.0 Ω)( 1.00× 10 −3 C 5.00× 10 −6 F)+(5.00 A)=25.00 A$

Conclusion:

Therefore, the value of current in battery is $25.00 A$ .

(b)

To determine

The value of resistances $R1, R2$ and $R3$ .

(b)

Expert Solution

Answer to Problem 100P

The value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Explanation of Solution

Formula used:

Apply loop rule to resistors $R3$ , $50.0 Ω$ , $5.00 Ω$ including battery,

$ε−(IR1)R1−(I50.0 Ω)(50.0 Ω)−(I5.00 Ω)(5.00 Ω)=0$ ..... (1)

Here, $IR1=Ibattery=25.0 A$

Apply loop rule to resistors $10.0 Ω$ , $50.0 Ω$ , $R2$ including battery,

$−(Ibattery−I5.00 Ω)(10.0 Ω)−IR2⋅R2+(I50.0 Ω)(50.0 Ω)=0$ ..... (2)

Apply loop rule to resistors $R1$ , $10.0 Ω$ , $R3$ including battery,

$ε−(IR1)R1−(Ibattery−I5.00 Ω)(10.0 Ω)−(I10.0 Ω−IR2)R3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R1$ is calculated as,

$(310 V)−(25.0 A)R1−(5.00 A)(50.0 Ω)−(10.0 A)(5.00 Ω)=0(25.0 A)R1=10 A⋅ΩR1=1025 ΩR1=0.40 Ω$

From equation (2), the resistance $R2$ is calculated as,

$−(25.0 A−5.00 A)(10.0 Ω)−(5.00 A)⋅R2+(5.00 A)(50.0 Ω)=0(5.00 A)⋅R2=50 A⋅ΩR2=10.0 Ω$

From equation (3), the resistance $R3$ is calculated as,

$(310 V)−(25 A)(0.40 Ω)−(25 A−5 A)(10.0 Ω)−(20 A−5 A)R3=0(100 A⋅Ω)−(20.0 A−5.00 A)R3=015R3=100 ΩR3=6.666 ΩR3≈6.67 Ω$

Conclusion:

Therefore, the value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

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Answer 2

Question

Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

(a)

Expert Solution

Answer to Problem 100P

The value of current in battery is $25.00 A$ .

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00 mC$ .

The value of capacitance is $C=5.00 μF$ .

The emf is $ε=310 V$ .

The current in $50.0 Ω$ resistor is $I50.0 Ω=5.00 A$ .

The current in $R2$ resistor is $IR2=5.00 A$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

$Ibattery=I10.0 Ω+I50.0 Ω$ ..... (1)

Calculation:

The expression for current in $10.0 Ω$ resistor is,

$I10.0 Ω=V 10.0 Ω10.0 Ω=1( 10.0 Ω)(qC) (∵V 10 Ω=qC)$ ..... (2)

From equation (1) and (2),

$Ibattery=1(10.0 Ω)(qC)+I50.0 Ω$

The current in battery is calculated as,

$Ibattery=1( 10.0 Ω)( 1.00 mC 5.00 μF)+(5.00 A)=1( 10.0 Ω)( 1.00× 10 −3 C 5.00× 10 −6 F)+(5.00 A)=25.00 A$

Conclusion:

Therefore, the value of current in battery is $25.00 A$ .

(b)

To determine

The value of resistances $R1, R2$ and $R3$ .

(b)

Expert Solution

Answer to Problem 100P

The value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Explanation of Solution

Formula used:

Apply loop rule to resistors $R3$ , $50.0 Ω$ , $5.00 Ω$ including battery,

$ε−(IR1)R1−(I50.0 Ω)(50.0 Ω)−(I5.00 Ω)(5.00 Ω)=0$ ..... (1)

Here, $IR1=Ibattery=25.0 A$

Apply loop rule to resistors $10.0 Ω$ , $50.0 Ω$ , $R2$ including battery,

$−(Ibattery−I5.00 Ω)(10.0 Ω)−IR2⋅R2+(I50.0 Ω)(50.0 Ω)=0$ ..... (2)

Apply loop rule to resistors $R1$ , $10.0 Ω$ , $R3$ including battery,

$ε−(IR1)R1−(Ibattery−I5.00 Ω)(10.0 Ω)−(I10.0 Ω−IR2)R3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R1$ is calculated as,

$(310 V)−(25.0 A)R1−(5.00 A)(50.0 Ω)−(10.0 A)(5.00 Ω)=0(25.0 A)R1=10 A⋅ΩR1=1025 ΩR1=0.40 Ω$

From equation (2), the resistance $R2$ is calculated as,

$−(25.0 A−5.00 A)(10.0 Ω)−(5.00 A)⋅R2+(5.00 A)(50.0 Ω)=0(5.00 A)⋅R2=50 A⋅ΩR2=10.0 Ω$

From equation (3), the resistance $R3$ is calculated as,

$(310 V)−(25 A)(0.40 Ω)−(25 A−5 A)(10.0 Ω)−(20 A−5 A)R3=0(100 A⋅Ω)−(20.0 A−5.00 A)R3=015R3=100 ΩR3=6.666 ΩR3≈6.67 Ω$

Conclusion:

Therefore, the value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

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Answer 3

Question

Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

(a)

Expert Solution

Answer to Problem 100P

The value of current in battery is $25.00 A$ .

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00 mC$ .

The value of capacitance is $C=5.00 μF$ .

The emf is $ε=310 V$ .

The current in $50.0 Ω$ resistor is $I50.0 Ω=5.00 A$ .

The current in $R2$ resistor is $IR2=5.00 A$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

$Ibattery=I10.0 Ω+I50.0 Ω$ ..... (1)

Calculation:

The expression for current in $10.0 Ω$ resistor is,

$I10.0 Ω=V 10.0 Ω10.0 Ω=1( 10.0 Ω)(qC) (∵V 10 Ω=qC)$ ..... (2)

From equation (1) and (2),

$Ibattery=1(10.0 Ω)(qC)+I50.0 Ω$

The current in battery is calculated as,

$Ibattery=1( 10.0 Ω)( 1.00 mC 5.00 μF)+(5.00 A)=1( 10.0 Ω)( 1.00× 10 −3 C 5.00× 10 −6 F)+(5.00 A)=25.00 A$

Conclusion:

Therefore, the value of current in battery is $25.00 A$ .

(b)

To determine

The value of resistances $R1, R2$ and $R3$ .

(b)

Expert Solution

Answer to Problem 100P

The value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Explanation of Solution

Formula used:

Apply loop rule to resistors $R3$ , $50.0 Ω$ , $5.00 Ω$ including battery,

$ε−(IR1)R1−(I50.0 Ω)(50.0 Ω)−(I5.00 Ω)(5.00 Ω)=0$ ..... (1)

Here, $IR1=Ibattery=25.0 A$

Apply loop rule to resistors $10.0 Ω$ , $50.0 Ω$ , $R2$ including battery,

$−(Ibattery−I5.00 Ω)(10.0 Ω)−IR2⋅R2+(I50.0 Ω)(50.0 Ω)=0$ ..... (2)

Apply loop rule to resistors $R1$ , $10.0 Ω$ , $R3$ including battery,

$ε−(IR1)R1−(Ibattery−I5.00 Ω)(10.0 Ω)−(I10.0 Ω−IR2)R3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R1$ is calculated as,

$(310 V)−(25.0 A)R1−(5.00 A)(50.0 Ω)−(10.0 A)(5.00 Ω)=0(25.0 A)R1=10 A⋅ΩR1=1025 ΩR1=0.40 Ω$

From equation (2), the resistance $R2$ is calculated as,

$−(25.0 A−5.00 A)(10.0 Ω)−(5.00 A)⋅R2+(5.00 A)(50.0 Ω)=0(5.00 A)⋅R2=50 A⋅ΩR2=10.0 Ω$

From equation (3), the resistance $R3$ is calculated as,

$(310 V)−(25 A)(0.40 Ω)−(25 A−5 A)(10.0 Ω)−(20 A−5 A)R3=0(100 A⋅Ω)−(20.0 A−5.00 A)R3=015R3=100 ΩR3=6.666 ΩR3≈6.67 Ω$

Conclusion:

Therefore, the value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

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Answer 4

Question

Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

(a)

Expert Solution

Answer to Problem 100P

The value of current in battery is $25.00 A$ .

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00 mC$ .

The value of capacitance is $C=5.00 μF$ .

The emf is $ε=310 V$ .

The current in $50.0 Ω$ resistor is $I50.0 Ω=5.00 A$ .

The current in $R2$ resistor is $IR2=5.00 A$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

$Ibattery=I10.0 Ω+I50.0 Ω$ ..... (1)

Calculation:

The expression for current in $10.0 Ω$ resistor is,

$I10.0 Ω=V 10.0 Ω10.0 Ω=1( 10.0 Ω)(qC) (∵V 10 Ω=qC)$ ..... (2)

From equation (1) and (2),

$Ibattery=1(10.0 Ω)(qC)+I50.0 Ω$

The current in battery is calculated as,

$Ibattery=1( 10.0 Ω)( 1.00 mC 5.00 μF)+(5.00 A)=1( 10.0 Ω)( 1.00× 10 −3 C 5.00× 10 −6 F)+(5.00 A)=25.00 A$

Conclusion:

Therefore, the value of current in battery is $25.00 A$ .

(b)

To determine

The value of resistances $R1, R2$ and $R3$ .

(b)

Expert Solution

Answer to Problem 100P

The value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Explanation of Solution

Formula used:

Apply loop rule to resistors $R3$ , $50.0 Ω$ , $5.00 Ω$ including battery,

$ε−(IR1)R1−(I50.0 Ω)(50.0 Ω)−(I5.00 Ω)(5.00 Ω)=0$ ..... (1)

Here, $IR1=Ibattery=25.0 A$

Apply loop rule to resistors $10.0 Ω$ , $50.0 Ω$ , $R2$ including battery,

$−(Ibattery−I5.00 Ω)(10.0 Ω)−IR2⋅R2+(I50.0 Ω)(50.0 Ω)=0$ ..... (2)

Apply loop rule to resistors $R1$ , $10.0 Ω$ , $R3$ including battery,

$ε−(IR1)R1−(Ibattery−I5.00 Ω)(10.0 Ω)−(I10.0 Ω−IR2)R3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R1$ is calculated as,

$(310 V)−(25.0 A)R1−(5.00 A)(50.0 Ω)−(10.0 A)(5.00 Ω)=0(25.0 A)R1=10 A⋅ΩR1=1025 ΩR1=0.40 Ω$

From equation (2), the resistance $R2$ is calculated as,

$−(25.0 A−5.00 A)(10.0 Ω)−(5.00 A)⋅R2+(5.00 A)(50.0 Ω)=0(5.00 A)⋅R2=50 A⋅ΩR2=10.0 Ω$

From equation (3), the resistance $R3$ is calculated as,

$(310 V)−(25 A)(0.40 Ω)−(25 A−5 A)(10.0 Ω)−(20 A−5 A)R3=0(100 A⋅Ω)−(20.0 A−5.00 A)R3=015R3=100 ΩR3=6.666 ΩR3≈6.67 Ω$

Conclusion:

Therefore, the value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

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Answer 5

Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

(a)

Expert Solution

Answer to Problem 100P

The value of current in battery is $25.00 A$ .

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00 mC$ .

The value of capacitance is $C=5.00 μF$ .

The emf is $ε=310 V$ .

The current in $50.0 Ω$ resistor is $I50.0 Ω=5.00 A$ .

The current in $R2$ resistor is $IR2=5.00 A$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

$Ibattery=I10.0 Ω+I50.0 Ω$ ..... (1)

Calculation:

The expression for current in $10.0 Ω$ resistor is,

$I10.0 Ω=V 10.0 Ω10.0 Ω=1( 10.0 Ω)(qC) (∵V 10 Ω=qC)$ ..... (2)

From equation (1) and (2),

$Ibattery=1(10.0 Ω)(qC)+I50.0 Ω$

The current in battery is calculated as,

$Ibattery=1( 10.0 Ω)( 1.00 mC 5.00 μF)+(5.00 A)=1( 10.0 Ω)( 1.00× 10 −3 C 5.00× 10 −6 F)+(5.00 A)=25.00 A$

Conclusion:

Therefore, the value of current in battery is $25.00 A$ .

(b)

To determine

The value of resistances $R1, R2$ and $R3$ .

(b)

Expert Solution

Answer to Problem 100P

The value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Explanation of Solution

Formula used:

Apply loop rule to resistors $R3$ , $50.0 Ω$ , $5.00 Ω$ including battery,

$ε−(IR1)R1−(I50.0 Ω)(50.0 Ω)−(I5.00 Ω)(5.00 Ω)=0$ ..... (1)

Here, $IR1=Ibattery=25.0 A$

Apply loop rule to resistors $10.0 Ω$ , $50.0 Ω$ , $R2$ including battery,

$−(Ibattery−I5.00 Ω)(10.0 Ω)−IR2⋅R2+(I50.0 Ω)(50.0 Ω)=0$ ..... (2)

Apply loop rule to resistors $R1$ , $10.0 Ω$ , $R3$ including battery,

$ε−(IR1)R1−(Ibattery−I5.00 Ω)(10.0 Ω)−(I10.0 Ω−IR2)R3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R1$ is calculated as,

$(310 V)−(25.0 A)R1−(5.00 A)(50.0 Ω)−(10.0 A)(5.00 Ω)=0(25.0 A)R1=10 A⋅ΩR1=1025 ΩR1=0.40 Ω$

From equation (2), the resistance $R2$ is calculated as,

$−(25.0 A−5.00 A)(10.0 Ω)−(5.00 A)⋅R2+(5.00 A)(50.0 Ω)=0(5.00 A)⋅R2=50 A⋅ΩR2=10.0 Ω$

From equation (3), the resistance $R3$ is calculated as,

$(310 V)−(25 A)(0.40 Ω)−(25 A−5 A)(10.0 Ω)−(20 A−5 A)R3=0(100 A⋅Ω)−(20.0 A−5.00 A)R3=015R3=100 ΩR3=6.666 ΩR3≈6.67 Ω$

Conclusion:

Therefore, the value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Answer 6

Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

(a)

Expert Solution

Answer to Problem 100P

The value of current in battery is $25.00 A$ .

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00 mC$ .

The value of capacitance is $C=5.00 μF$ .

The emf is $ε=310 V$ .

The current in $50.0 Ω$ resistor is $I50.0 Ω=5.00 A$ .

The current in $R2$ resistor is $IR2=5.00 A$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

$Ibattery=I10.0 Ω+I50.0 Ω$ ..... (1)

Calculation:

The expression for current in $10.0 Ω$ resistor is,

$I10.0 Ω=V 10.0 Ω10.0 Ω=1( 10.0 Ω)(qC) (∵V 10 Ω=qC)$ ..... (2)

From equation (1) and (2),

$Ibattery=1(10.0 Ω)(qC)+I50.0 Ω$

The current in battery is calculated as,

$Ibattery=1( 10.0 Ω)( 1.00 mC 5.00 μF)+(5.00 A)=1( 10.0 Ω)( 1.00× 10 −3 C 5.00× 10 −6 F)+(5.00 A)=25.00 A$

Conclusion:

Therefore, the value of current in battery is $25.00 A$ .

Answer 7

Chapter 25, Problem 100P

(a)

To determine

The value of current in battery.

Answer 8

(a)

Expert Solution

Answer to Problem 100P

The value of current in battery is $25.00 A$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The value of charge on capacitor is $q=1.00 mC$ .

The value of capacitance is $C=5.00 μF$ .

The emf is $ε=310 V$ .

The current in $50.0 Ω$ resistor is $I50.0 Ω=5.00 A$ .

The current in $R2$ resistor is $IR2=5.00 A$ .

Formula used:

By the Kirchhoff’s law, the expression for the current in battery is given as,

$Ibattery=I10.0 Ω+I50.0 Ω$ ..... (1)

Calculation:

The expression for current in $10.0 Ω$ resistor is,

$I10.0 Ω=V 10.0 Ω10.0 Ω=1( 10.0 Ω)(qC) (∵V 10 Ω=qC)$ ..... (2)

From equation (1) and (2),

$Ibattery=1(10.0 Ω)(qC)+I50.0 Ω$

The current in battery is calculated as,

$Ibattery=1( 10.0 Ω)( 1.00 mC 5.00 μF)+(5.00 A)=1( 10.0 Ω)( 1.00× 10 −3 C 5.00× 10 −6 F)+(5.00 A)=25.00 A$

Conclusion:

Therefore, the value of current in battery is $25.00 A$ .

Answer 13

(b)

To determine

The value of resistances $R1, R2$ and $R3$ .

(b)

Expert Solution

Answer to Problem 100P

The value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Explanation of Solution

Formula used:

Apply loop rule to resistors $R3$ , $50.0 Ω$ , $5.00 Ω$ including battery,

$ε−(IR1)R1−(I50.0 Ω)(50.0 Ω)−(I5.00 Ω)(5.00 Ω)=0$ ..... (1)

Here, $IR1=Ibattery=25.0 A$

Apply loop rule to resistors $10.0 Ω$ , $50.0 Ω$ , $R2$ including battery,

$−(Ibattery−I5.00 Ω)(10.0 Ω)−IR2⋅R2+(I50.0 Ω)(50.0 Ω)=0$ ..... (2)

Apply loop rule to resistors $R1$ , $10.0 Ω$ , $R3$ including battery,

$ε−(IR1)R1−(Ibattery−I5.00 Ω)(10.0 Ω)−(I10.0 Ω−IR2)R3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R1$ is calculated as,

$(310 V)−(25.0 A)R1−(5.00 A)(50.0 Ω)−(10.0 A)(5.00 Ω)=0(25.0 A)R1=10 A⋅ΩR1=1025 ΩR1=0.40 Ω$

From equation (2), the resistance $R2$ is calculated as,

$−(25.0 A−5.00 A)(10.0 Ω)−(5.00 A)⋅R2+(5.00 A)(50.0 Ω)=0(5.00 A)⋅R2=50 A⋅ΩR2=10.0 Ω$

From equation (3), the resistance $R3$ is calculated as,

$(310 V)−(25 A)(0.40 Ω)−(25 A−5 A)(10.0 Ω)−(20 A−5 A)R3=0(100 A⋅Ω)−(20.0 A−5.00 A)R3=015R3=100 ΩR3=6.666 ΩR3≈6.67 Ω$

Conclusion:

Therefore, the value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Answer 14

(b)

To determine

The value of resistances $R1, R2$ and $R3$ .

Answer 15

(b)

Expert Solution

Answer to Problem 100P

The value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

Apply loop rule to resistors $R3$ , $50.0 Ω$ , $5.00 Ω$ including battery,

$ε−(IR1)R1−(I50.0 Ω)(50.0 Ω)−(I5.00 Ω)(5.00 Ω)=0$ ..... (1)

Here, $IR1=Ibattery=25.0 A$

Apply loop rule to resistors $10.0 Ω$ , $50.0 Ω$ , $R2$ including battery,

$−(Ibattery−I5.00 Ω)(10.0 Ω)−IR2⋅R2+(I50.0 Ω)(50.0 Ω)=0$ ..... (2)

Apply loop rule to resistors $R1$ , $10.0 Ω$ , $R3$ including battery,

$ε−(IR1)R1−(Ibattery−I5.00 Ω)(10.0 Ω)−(I10.0 Ω−IR2)R3=0$ ..... (3)

Calculation:

From equation (1), the resistance $R1$ is calculated as,

$(310 V)−(25.0 A)R1−(5.00 A)(50.0 Ω)−(10.0 A)(5.00 Ω)=0(25.0 A)R1=10 A⋅ΩR1=1025 ΩR1=0.40 Ω$

From equation (2), the resistance $R2$ is calculated as,

$−(25.0 A−5.00 A)(10.0 Ω)−(5.00 A)⋅R2+(5.00 A)(50.0 Ω)=0(5.00 A)⋅R2=50 A⋅ΩR2=10.0 Ω$

From equation (3), the resistance $R3$ is calculated as,

$(310 V)−(25 A)(0.40 Ω)−(25 A−5 A)(10.0 Ω)−(20 A−5 A)R3=0(100 A⋅Ω)−(20.0 A−5.00 A)R3=015R3=100 ΩR3=6.666 ΩR3≈6.67 Ω$

Conclusion:

Therefore, the value of resistances $R1, R2$ and $R3$ are $0.40 Ω$ , $10.0 Ω$ and $6.67 Ω$ respectively.

Answer 20

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

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The value of current in battery.

Answer to Problem 100P

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The value of resistances $R1, R2$ and $R3$ .

Answer to Problem 100P

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The value of current in battery.

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The value of current in battery.

Answer to Problem 100P

Explanation of Solution

The value of resistances R1, R2<m:math><m:mrow><m:msub><m:mi>R</m:mi><m:mn>1</m:mn></m:msub><m:mo>,</m:mo><m:mtext> </m:mtext><m:msub><m:mi>R</m:mi><m:mn>2</m:mn></m:msub></m:mrow></m:math> and R3<m:math><m:mrow><m:msub><m:mi>R</m:mi><m:mn>3</m:mn></m:msub></m:mrow></m:math> .

Answer to Problem 100P

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Chapter 25 Solutions

The value of resistances $R1, R2$ and $R3$ .