The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each branch of the circuit. (b) Find the energy delivered by each battery. (c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. (e) Find the total amount of energy transformed into internal energy in the resistors. Figure P21.47 Problems 47 and 48.

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Answer 1

Textbook Question

Chapter 21, Problem 47P

The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each branch of the circuit. (b) Find the energy delivered by each battery. (c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. (e) Find the total amount of energy transformed into internal energy in the resistors.

Figure P21.47 Problems 47 and 48.

Chapter 21, Problem 47P, The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each

(a)

Expert Solution

To determine

The current in each branch of the circuit.

Answer to Problem 47P

The current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 47P

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$ .

$12.0 V−(4.00 Ω)I3−(6.00 Ω)I2−4.00 V=0(4.00 Ω)I3+(6.00 Ω)I2=8.00 V$ (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$ .

$−(6.00 Ω)I2−4.00 V+(8.00 Ω)I1=0(6.00 Ω)I2+4.00 V=(8.00 Ω)I1$ (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$ .

$I3=I1+I2$ (III)

Use equation (III) in (I) and solve the equation.

$(4.00 Ω)(I1+I2)+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(4.00 Ω)I2+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(10.0 Ω)I2=8.00 V$ (IV)

Use equation (II) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)$ (V)

Use equation (IV) in (V) to solve for $I1$ .

$(4.00 Ω)I1+(10.0 Ω)[(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)]=8.00 V$ (VI)

Use equation (VI) to solve for $I1$ .

$I1=352 Ω(8.00 V+203 V)=0.846 A$ (VII)

Use equation (VII) in (V) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)(0.846 A)−(2.00 V3.00 Ω)=0.462 A$ (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I3$ .

$I3=(0.846 A)+(0.462 A)=1.31 A$ (IX)

Conclusion:

Therefore, the current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

(b)

Expert Solution

To determine

The energy delivered to each battery.

Answer to Problem 47P

The energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Explanation of Solution

Write the expression for the energy delivered to the battery.

$ΔUB=PΔt$ (X)

Here, $ΔUB$ is the energy delivered, $P$ is the power, $Δt$ is the time interval.

Write the expression for the $P$ .

$P=I(ΔV)$ (XI)

Here, $ΔV$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $ΔU$ .

$ΔUB=(ΔV)I(Δt)$ (XII)

Conclusion:

Substitute $4.00 V$ for $ΔV$ , $−0.462 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $4.00 V$ battery.

$ΔUB=(4.00 V)(−0.462 A)(2.0 min×60 s1 min)=−222 J$

Substitute $12.0 V$ for $ΔV$ , $1.31 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $12.0 V$ battery.

$ΔUB=(12.0 V)(1.31 A)(2.0 min×60 s1 min)=1.8×103 J=1.8 kJ$

Therefore, the energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

(c)

Expert Solution

To determine

The energy delivered to each resistor.

Answer to Problem 47P

The energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Explanation of Solution

Write the expression for the energy delivered to the resistor.

$ΔUR=I2RΔt$ (XIII)

Here, $ΔUR$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846 A$ for $I$ , $8.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $8.00 Ω$ resistor.

$ΔUR=(0.846 A)2(8.00 Ω)(2.0 min×60 s1 min)=687 J$

Substitute $0.462 A$ for $I$ , $5.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $5.00 Ω$ resistor.

$ΔUR=(0.462 A)2(5.00 Ω)(2.0 min×60 s1 min)=128 J$

Substitute $0.462 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor.

$ΔUR=(0.462 A)2(1.00 Ω)(2.0 min×60 s1 min)=25.6 J$

Substitute $1.31 A$ for $I$ , $3.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $3.00 Ω$ resistor.

$ΔUR=(1.31 A)2(3.00 Ω)(2.0 min×60 s1 min)=616 J$

Substitute $1.31 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor in the right hand branch.

$ΔUR=(1.31 A)2(1.00 Ω)(2.0 min×60 s1 min)=205 J$

Therefore, the energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

(d)

Expert Solution

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Explanation of Solution

The chemical energy in the $12.0 V$ battery is transformed into internal energy in the resistors. The $4.00 V$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0 V$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

(e)

Expert Solution

To determine

The total amount of energy transformed into internal energy in the resistors.

Answer to Problem 47P

The total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

$ΔUT=ΔUB1+ΔUB2$ (XIV)

Here, $ΔUT$ is the total amount of energy transformed, $ΔUB1$ is the energy delivered to $4.00 V$ battery, $ΔUB2$ is the energy delivered to $12.0 V$ battery.

Conclusion:

Substitute $−222 J$ for $ΔUB1$ , $1.88 kJ$ for $ΔUB2$ in equation (XIV) to find $ΔUT$ .

$ΔUT=−222 J+(1.88 kJ×103J1 kJ)=1.66×103J=1.66 kJ$

Therefore, the total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

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Answer 2

Textbook Question

Chapter 21, Problem 47P

The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each branch of the circuit. (b) Find the energy delivered by each battery. (c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. (e) Find the total amount of energy transformed into internal energy in the resistors.

Figure P21.47 Problems 47 and 48.

Chapter 21, Problem 47P, The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each

(a)

Expert Solution

To determine

The current in each branch of the circuit.

Answer to Problem 47P

The current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 47P

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$ .

$12.0 V−(4.00 Ω)I3−(6.00 Ω)I2−4.00 V=0(4.00 Ω)I3+(6.00 Ω)I2=8.00 V$ (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$ .

$−(6.00 Ω)I2−4.00 V+(8.00 Ω)I1=0(6.00 Ω)I2+4.00 V=(8.00 Ω)I1$ (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$ .

$I3=I1+I2$ (III)

Use equation (III) in (I) and solve the equation.

$(4.00 Ω)(I1+I2)+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(4.00 Ω)I2+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(10.0 Ω)I2=8.00 V$ (IV)

Use equation (II) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)$ (V)

Use equation (IV) in (V) to solve for $I1$ .

$(4.00 Ω)I1+(10.0 Ω)[(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)]=8.00 V$ (VI)

Use equation (VI) to solve for $I1$ .

$I1=352 Ω(8.00 V+203 V)=0.846 A$ (VII)

Use equation (VII) in (V) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)(0.846 A)−(2.00 V3.00 Ω)=0.462 A$ (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I3$ .

$I3=(0.846 A)+(0.462 A)=1.31 A$ (IX)

Conclusion:

Therefore, the current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

(b)

Expert Solution

To determine

The energy delivered to each battery.

Answer to Problem 47P

The energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Explanation of Solution

Write the expression for the energy delivered to the battery.

$ΔUB=PΔt$ (X)

Here, $ΔUB$ is the energy delivered, $P$ is the power, $Δt$ is the time interval.

Write the expression for the $P$ .

$P=I(ΔV)$ (XI)

Here, $ΔV$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $ΔU$ .

$ΔUB=(ΔV)I(Δt)$ (XII)

Conclusion:

Substitute $4.00 V$ for $ΔV$ , $−0.462 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $4.00 V$ battery.

$ΔUB=(4.00 V)(−0.462 A)(2.0 min×60 s1 min)=−222 J$

Substitute $12.0 V$ for $ΔV$ , $1.31 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $12.0 V$ battery.

$ΔUB=(12.0 V)(1.31 A)(2.0 min×60 s1 min)=1.8×103 J=1.8 kJ$

Therefore, the energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

(c)

Expert Solution

To determine

The energy delivered to each resistor.

Answer to Problem 47P

The energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Explanation of Solution

Write the expression for the energy delivered to the resistor.

$ΔUR=I2RΔt$ (XIII)

Here, $ΔUR$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846 A$ for $I$ , $8.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $8.00 Ω$ resistor.

$ΔUR=(0.846 A)2(8.00 Ω)(2.0 min×60 s1 min)=687 J$

Substitute $0.462 A$ for $I$ , $5.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $5.00 Ω$ resistor.

$ΔUR=(0.462 A)2(5.00 Ω)(2.0 min×60 s1 min)=128 J$

Substitute $0.462 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor.

$ΔUR=(0.462 A)2(1.00 Ω)(2.0 min×60 s1 min)=25.6 J$

Substitute $1.31 A$ for $I$ , $3.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $3.00 Ω$ resistor.

$ΔUR=(1.31 A)2(3.00 Ω)(2.0 min×60 s1 min)=616 J$

Substitute $1.31 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor in the right hand branch.

$ΔUR=(1.31 A)2(1.00 Ω)(2.0 min×60 s1 min)=205 J$

Therefore, the energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

(d)

Expert Solution

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Explanation of Solution

The chemical energy in the $12.0 V$ battery is transformed into internal energy in the resistors. The $4.00 V$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0 V$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

(e)

Expert Solution

To determine

The total amount of energy transformed into internal energy in the resistors.

Answer to Problem 47P

The total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

$ΔUT=ΔUB1+ΔUB2$ (XIV)

Here, $ΔUT$ is the total amount of energy transformed, $ΔUB1$ is the energy delivered to $4.00 V$ battery, $ΔUB2$ is the energy delivered to $12.0 V$ battery.

Conclusion:

Substitute $−222 J$ for $ΔUB1$ , $1.88 kJ$ for $ΔUB2$ in equation (XIV) to find $ΔUT$ .

$ΔUT=−222 J+(1.88 kJ×103J1 kJ)=1.66×103J=1.66 kJ$

Therefore, the total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

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Answer 3

Textbook Question

Chapter 21, Problem 47P

The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each branch of the circuit. (b) Find the energy delivered by each battery. (c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. (e) Find the total amount of energy transformed into internal energy in the resistors.

Figure P21.47 Problems 47 and 48.

Chapter 21, Problem 47P, The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each

(a)

Expert Solution

To determine

The current in each branch of the circuit.

Answer to Problem 47P

The current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 47P

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$ .

$12.0 V−(4.00 Ω)I3−(6.00 Ω)I2−4.00 V=0(4.00 Ω)I3+(6.00 Ω)I2=8.00 V$ (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$ .

$−(6.00 Ω)I2−4.00 V+(8.00 Ω)I1=0(6.00 Ω)I2+4.00 V=(8.00 Ω)I1$ (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$ .

$I3=I1+I2$ (III)

Use equation (III) in (I) and solve the equation.

$(4.00 Ω)(I1+I2)+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(4.00 Ω)I2+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(10.0 Ω)I2=8.00 V$ (IV)

Use equation (II) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)$ (V)

Use equation (IV) in (V) to solve for $I1$ .

$(4.00 Ω)I1+(10.0 Ω)[(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)]=8.00 V$ (VI)

Use equation (VI) to solve for $I1$ .

$I1=352 Ω(8.00 V+203 V)=0.846 A$ (VII)

Use equation (VII) in (V) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)(0.846 A)−(2.00 V3.00 Ω)=0.462 A$ (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I3$ .

$I3=(0.846 A)+(0.462 A)=1.31 A$ (IX)

Conclusion:

Therefore, the current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

(b)

Expert Solution

To determine

The energy delivered to each battery.

Answer to Problem 47P

The energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Explanation of Solution

Write the expression for the energy delivered to the battery.

$ΔUB=PΔt$ (X)

Here, $ΔUB$ is the energy delivered, $P$ is the power, $Δt$ is the time interval.

Write the expression for the $P$ .

$P=I(ΔV)$ (XI)

Here, $ΔV$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $ΔU$ .

$ΔUB=(ΔV)I(Δt)$ (XII)

Conclusion:

Substitute $4.00 V$ for $ΔV$ , $−0.462 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $4.00 V$ battery.

$ΔUB=(4.00 V)(−0.462 A)(2.0 min×60 s1 min)=−222 J$

Substitute $12.0 V$ for $ΔV$ , $1.31 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $12.0 V$ battery.

$ΔUB=(12.0 V)(1.31 A)(2.0 min×60 s1 min)=1.8×103 J=1.8 kJ$

Therefore, the energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

(c)

Expert Solution

To determine

The energy delivered to each resistor.

Answer to Problem 47P

The energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Explanation of Solution

Write the expression for the energy delivered to the resistor.

$ΔUR=I2RΔt$ (XIII)

Here, $ΔUR$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846 A$ for $I$ , $8.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $8.00 Ω$ resistor.

$ΔUR=(0.846 A)2(8.00 Ω)(2.0 min×60 s1 min)=687 J$

Substitute $0.462 A$ for $I$ , $5.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $5.00 Ω$ resistor.

$ΔUR=(0.462 A)2(5.00 Ω)(2.0 min×60 s1 min)=128 J$

Substitute $0.462 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor.

$ΔUR=(0.462 A)2(1.00 Ω)(2.0 min×60 s1 min)=25.6 J$

Substitute $1.31 A$ for $I$ , $3.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $3.00 Ω$ resistor.

$ΔUR=(1.31 A)2(3.00 Ω)(2.0 min×60 s1 min)=616 J$

Substitute $1.31 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor in the right hand branch.

$ΔUR=(1.31 A)2(1.00 Ω)(2.0 min×60 s1 min)=205 J$

Therefore, the energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

(d)

Expert Solution

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Explanation of Solution

The chemical energy in the $12.0 V$ battery is transformed into internal energy in the resistors. The $4.00 V$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0 V$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

(e)

Expert Solution

To determine

The total amount of energy transformed into internal energy in the resistors.

Answer to Problem 47P

The total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

$ΔUT=ΔUB1+ΔUB2$ (XIV)

Here, $ΔUT$ is the total amount of energy transformed, $ΔUB1$ is the energy delivered to $4.00 V$ battery, $ΔUB2$ is the energy delivered to $12.0 V$ battery.

Conclusion:

Substitute $−222 J$ for $ΔUB1$ , $1.88 kJ$ for $ΔUB2$ in equation (XIV) to find $ΔUT$ .

$ΔUT=−222 J+(1.88 kJ×103J1 kJ)=1.66×103J=1.66 kJ$

Therefore, the total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

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Answer 4

Textbook Question

Chapter 21, Problem 47P

The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each branch of the circuit. (b) Find the energy delivered by each battery. (c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. (e) Find the total amount of energy transformed into internal energy in the resistors.

Figure P21.47 Problems 47 and 48.

Chapter 21, Problem 47P, The circuit shown in Figure P21.47 is connected for 2.00 min. (a) Determine the current in each

(a)

Expert Solution

To determine

The current in each branch of the circuit.

Answer to Problem 47P

The current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 47P

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$ .

$12.0 V−(4.00 Ω)I3−(6.00 Ω)I2−4.00 V=0(4.00 Ω)I3+(6.00 Ω)I2=8.00 V$ (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$ .

$−(6.00 Ω)I2−4.00 V+(8.00 Ω)I1=0(6.00 Ω)I2+4.00 V=(8.00 Ω)I1$ (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$ .

$I3=I1+I2$ (III)

Use equation (III) in (I) and solve the equation.

$(4.00 Ω)(I1+I2)+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(4.00 Ω)I2+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(10.0 Ω)I2=8.00 V$ (IV)

Use equation (II) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)$ (V)

Use equation (IV) in (V) to solve for $I1$ .

$(4.00 Ω)I1+(10.0 Ω)[(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)]=8.00 V$ (VI)

Use equation (VI) to solve for $I1$ .

$I1=352 Ω(8.00 V+203 V)=0.846 A$ (VII)

Use equation (VII) in (V) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)(0.846 A)−(2.00 V3.00 Ω)=0.462 A$ (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I3$ .

$I3=(0.846 A)+(0.462 A)=1.31 A$ (IX)

Conclusion:

Therefore, the current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

(b)

Expert Solution

To determine

The energy delivered to each battery.

Answer to Problem 47P

The energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Explanation of Solution

Write the expression for the energy delivered to the battery.

$ΔUB=PΔt$ (X)

Here, $ΔUB$ is the energy delivered, $P$ is the power, $Δt$ is the time interval.

Write the expression for the $P$ .

$P=I(ΔV)$ (XI)

Here, $ΔV$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $ΔU$ .

$ΔUB=(ΔV)I(Δt)$ (XII)

Conclusion:

Substitute $4.00 V$ for $ΔV$ , $−0.462 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $4.00 V$ battery.

$ΔUB=(4.00 V)(−0.462 A)(2.0 min×60 s1 min)=−222 J$

Substitute $12.0 V$ for $ΔV$ , $1.31 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $12.0 V$ battery.

$ΔUB=(12.0 V)(1.31 A)(2.0 min×60 s1 min)=1.8×103 J=1.8 kJ$

Therefore, the energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

(c)

Expert Solution

To determine

The energy delivered to each resistor.

Answer to Problem 47P

The energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Explanation of Solution

Write the expression for the energy delivered to the resistor.

$ΔUR=I2RΔt$ (XIII)

Here, $ΔUR$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846 A$ for $I$ , $8.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $8.00 Ω$ resistor.

$ΔUR=(0.846 A)2(8.00 Ω)(2.0 min×60 s1 min)=687 J$

Substitute $0.462 A$ for $I$ , $5.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $5.00 Ω$ resistor.

$ΔUR=(0.462 A)2(5.00 Ω)(2.0 min×60 s1 min)=128 J$

Substitute $0.462 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor.

$ΔUR=(0.462 A)2(1.00 Ω)(2.0 min×60 s1 min)=25.6 J$

Substitute $1.31 A$ for $I$ , $3.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $3.00 Ω$ resistor.

$ΔUR=(1.31 A)2(3.00 Ω)(2.0 min×60 s1 min)=616 J$

Substitute $1.31 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor in the right hand branch.

$ΔUR=(1.31 A)2(1.00 Ω)(2.0 min×60 s1 min)=205 J$

Therefore, the energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

(d)

Expert Solution

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Explanation of Solution

The chemical energy in the $12.0 V$ battery is transformed into internal energy in the resistors. The $4.00 V$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0 V$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

(e)

Expert Solution

To determine

The total amount of energy transformed into internal energy in the resistors.

Answer to Problem 47P

The total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

$ΔUT=ΔUB1+ΔUB2$ (XIV)

Here, $ΔUT$ is the total amount of energy transformed, $ΔUB1$ is the energy delivered to $4.00 V$ battery, $ΔUB2$ is the energy delivered to $12.0 V$ battery.

Conclusion:

Substitute $−222 J$ for $ΔUB1$ , $1.88 kJ$ for $ΔUB2$ in equation (XIV) to find $ΔUT$ .

$ΔUT=−222 J+(1.88 kJ×103J1 kJ)=1.66×103J=1.66 kJ$

Therefore, the total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The current in each branch of the circuit.

Answer to Problem 47P

The current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 47P

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$ .

$12.0 V−(4.00 Ω)I3−(6.00 Ω)I2−4.00 V=0(4.00 Ω)I3+(6.00 Ω)I2=8.00 V$ (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$ .

$−(6.00 Ω)I2−4.00 V+(8.00 Ω)I1=0(6.00 Ω)I2+4.00 V=(8.00 Ω)I1$ (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$ .

$I3=I1+I2$ (III)

Use equation (III) in (I) and solve the equation.

$(4.00 Ω)(I1+I2)+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(4.00 Ω)I2+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(10.0 Ω)I2=8.00 V$ (IV)

Use equation (II) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)$ (V)

Use equation (IV) in (V) to solve for $I1$ .

$(4.00 Ω)I1+(10.0 Ω)[(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)]=8.00 V$ (VI)

Use equation (VI) to solve for $I1$ .

$I1=352 Ω(8.00 V+203 V)=0.846 A$ (VII)

Use equation (VII) in (V) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)(0.846 A)−(2.00 V3.00 Ω)=0.462 A$ (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I3$ .

$I3=(0.846 A)+(0.462 A)=1.31 A$ (IX)

Conclusion:

Therefore, the current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

(b)

Expert Solution

To determine

The energy delivered to each battery.

Answer to Problem 47P

The energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Explanation of Solution

Write the expression for the energy delivered to the battery.

$ΔUB=PΔt$ (X)

Here, $ΔUB$ is the energy delivered, $P$ is the power, $Δt$ is the time interval.

Write the expression for the $P$ .

$P=I(ΔV)$ (XI)

Here, $ΔV$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $ΔU$ .

$ΔUB=(ΔV)I(Δt)$ (XII)

Conclusion:

Substitute $4.00 V$ for $ΔV$ , $−0.462 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $4.00 V$ battery.

$ΔUB=(4.00 V)(−0.462 A)(2.0 min×60 s1 min)=−222 J$

Substitute $12.0 V$ for $ΔV$ , $1.31 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $12.0 V$ battery.

$ΔUB=(12.0 V)(1.31 A)(2.0 min×60 s1 min)=1.8×103 J=1.8 kJ$

Therefore, the energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

(c)

Expert Solution

To determine

The energy delivered to each resistor.

Answer to Problem 47P

The energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Explanation of Solution

Write the expression for the energy delivered to the resistor.

$ΔUR=I2RΔt$ (XIII)

Here, $ΔUR$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846 A$ for $I$ , $8.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $8.00 Ω$ resistor.

$ΔUR=(0.846 A)2(8.00 Ω)(2.0 min×60 s1 min)=687 J$

Substitute $0.462 A$ for $I$ , $5.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $5.00 Ω$ resistor.

$ΔUR=(0.462 A)2(5.00 Ω)(2.0 min×60 s1 min)=128 J$

Substitute $0.462 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor.

$ΔUR=(0.462 A)2(1.00 Ω)(2.0 min×60 s1 min)=25.6 J$

Substitute $1.31 A$ for $I$ , $3.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $3.00 Ω$ resistor.

$ΔUR=(1.31 A)2(3.00 Ω)(2.0 min×60 s1 min)=616 J$

Substitute $1.31 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor in the right hand branch.

$ΔUR=(1.31 A)2(1.00 Ω)(2.0 min×60 s1 min)=205 J$

Therefore, the energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

(d)

Expert Solution

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Explanation of Solution

The chemical energy in the $12.0 V$ battery is transformed into internal energy in the resistors. The $4.00 V$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0 V$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

(e)

Expert Solution

To determine

The total amount of energy transformed into internal energy in the resistors.

Answer to Problem 47P

The total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

$ΔUT=ΔUB1+ΔUB2$ (XIV)

Here, $ΔUT$ is the total amount of energy transformed, $ΔUB1$ is the energy delivered to $4.00 V$ battery, $ΔUB2$ is the energy delivered to $12.0 V$ battery.

Conclusion:

Substitute $−222 J$ for $ΔUB1$ , $1.88 kJ$ for $ΔUB2$ in equation (XIV) to find $ΔUT$ .

$ΔUT=−222 J+(1.88 kJ×103J1 kJ)=1.66×103J=1.66 kJ$

Therefore, the total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Answer 8

(a)

Expert Solution

To determine

The current in each branch of the circuit.

Answer to Problem 47P

The current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 47P

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$ .

$12.0 V−(4.00 Ω)I3−(6.00 Ω)I2−4.00 V=0(4.00 Ω)I3+(6.00 Ω)I2=8.00 V$ (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$ .

$−(6.00 Ω)I2−4.00 V+(8.00 Ω)I1=0(6.00 Ω)I2+4.00 V=(8.00 Ω)I1$ (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$ .

$I3=I1+I2$ (III)

Use equation (III) in (I) and solve the equation.

$(4.00 Ω)(I1+I2)+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(4.00 Ω)I2+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(10.0 Ω)I2=8.00 V$ (IV)

Use equation (II) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)$ (V)

Use equation (IV) in (V) to solve for $I1$ .

$(4.00 Ω)I1+(10.0 Ω)[(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)]=8.00 V$ (VI)

Use equation (VI) to solve for $I1$ .

$I1=352 Ω(8.00 V+203 V)=0.846 A$ (VII)

Use equation (VII) in (V) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)(0.846 A)−(2.00 V3.00 Ω)=0.462 A$ (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I3$ .

$I3=(0.846 A)+(0.462 A)=1.31 A$ (IX)

Conclusion:

Therefore, the current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The current in each branch of the circuit.

Answer 13

Answer to Problem 47P

The current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Answer 14

Explanation of Solution

Principles of Physics: A Calculus-Based Text, Chapter 21, Problem 47P

Write the expression for the Kirchhoff’s loop rule in second loop from the left of the figure $1$ .

$12.0 V−(4.00 Ω)I3−(6.00 Ω)I2−4.00 V=0(4.00 Ω)I3+(6.00 Ω)I2=8.00 V$ (I)

Here, $I$ is the current.

Write the expression for the Kirchhoff’s loop rule in first loop from the left of the figure $1$ .

$−(6.00 Ω)I2−4.00 V+(8.00 Ω)I1=0(6.00 Ω)I2+4.00 V=(8.00 Ω)I1$ (II)

Write the expression for the Kirchhoff’s junction rule in the figure $1$ .

$I3=I1+I2$ (III)

Use equation (III) in (I) and solve the equation.

$(4.00 Ω)(I1+I2)+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(4.00 Ω)I2+(6.00 Ω)I2=8.00 V(4.00 Ω)I1+(10.0 Ω)I2=8.00 V$ (IV)

Use equation (II) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)$ (V)

Use equation (IV) in (V) to solve for $I1$ .

$(4.00 Ω)I1+(10.0 Ω)[(4.00 Ω3.00 Ω)I1−(2.00 V3.00 Ω)]=8.00 V$ (VI)

Use equation (VI) to solve for $I1$ .

$I1=352 Ω(8.00 V+203 V)=0.846 A$ (VII)

Use equation (VII) in (V) to solve for $I2$ .

$I2=(4.00 Ω3.00 Ω)(0.846 A)−(2.00 V3.00 Ω)=0.462 A$ (VIII)

Use equation (VIII) and (VII)in (III) to solve for $I3$ .

$I3=(0.846 A)+(0.462 A)=1.31 A$ (IX)

Conclusion:

Therefore, the current $I1$ is $0.846 A down in the 8.00 Ω resistor_$ , $I2$ is $0.462 A down in the middle branch_$ and $I3$ is $1.31 A up in the right-hand branch_$ in the circuit.

Answer 15

(b)

Expert Solution

To determine

The energy delivered to each battery.

Answer to Problem 47P

The energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Explanation of Solution

Write the expression for the energy delivered to the battery.

$ΔUB=PΔt$ (X)

Here, $ΔUB$ is the energy delivered, $P$ is the power, $Δt$ is the time interval.

Write the expression for the $P$ .

$P=I(ΔV)$ (XI)

Here, $ΔV$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $ΔU$ .

$ΔUB=(ΔV)I(Δt)$ (XII)

Conclusion:

Substitute $4.00 V$ for $ΔV$ , $−0.462 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $4.00 V$ battery.

$ΔUB=(4.00 V)(−0.462 A)(2.0 min×60 s1 min)=−222 J$

Substitute $12.0 V$ for $ΔV$ , $1.31 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $12.0 V$ battery.

$ΔUB=(12.0 V)(1.31 A)(2.0 min×60 s1 min)=1.8×103 J=1.8 kJ$

Therefore, the energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The energy delivered to each battery.

Answer 20

Answer to Problem 47P

The energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Answer 21

Explanation of Solution

Write the expression for the energy delivered to the battery.

$ΔUB=PΔt$ (X)

Here, $ΔUB$ is the energy delivered, $P$ is the power, $Δt$ is the time interval.

Write the expression for the $P$ .

$P=I(ΔV)$ (XI)

Here, $ΔV$ is the voltage of the battery.

Use equation (XI) in (X) to solve for $ΔU$ .

$ΔUB=(ΔV)I(Δt)$ (XII)

Conclusion:

Substitute $4.00 V$ for $ΔV$ , $−0.462 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $4.00 V$ battery.

$ΔUB=(4.00 V)(−0.462 A)(2.0 min×60 s1 min)=−222 J$

Substitute $12.0 V$ for $ΔV$ , $1.31 A$ for $I$ , $2.00 min$ for $Δt$ in equation (XII) to find the power delivered to $12.0 V$ battery.

$ΔUB=(12.0 V)(1.31 A)(2.0 min×60 s1 min)=1.8×103 J=1.8 kJ$

Therefore, the energy delivered to $4.00 V$ battery is $−222 J_$ and $12.0 V$ battery is $1.88 kJ_$ .

Answer 22

(c)

Expert Solution

To determine

The energy delivered to each resistor.

Answer to Problem 47P

The energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Explanation of Solution

Write the expression for the energy delivered to the resistor.

$ΔUR=I2RΔt$ (XIII)

Here, $ΔUR$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846 A$ for $I$ , $8.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $8.00 Ω$ resistor.

$ΔUR=(0.846 A)2(8.00 Ω)(2.0 min×60 s1 min)=687 J$

Substitute $0.462 A$ for $I$ , $5.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $5.00 Ω$ resistor.

$ΔUR=(0.462 A)2(5.00 Ω)(2.0 min×60 s1 min)=128 J$

Substitute $0.462 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor.

$ΔUR=(0.462 A)2(1.00 Ω)(2.0 min×60 s1 min)=25.6 J$

Substitute $1.31 A$ for $I$ , $3.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $3.00 Ω$ resistor.

$ΔUR=(1.31 A)2(3.00 Ω)(2.0 min×60 s1 min)=616 J$

Substitute $1.31 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor in the right hand branch.

$ΔUR=(1.31 A)2(1.00 Ω)(2.0 min×60 s1 min)=205 J$

Therefore, the energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The energy delivered to each resistor.

Answer 27

Answer to Problem 47P

The energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Answer 28

Explanation of Solution

Write the expression for the energy delivered to the resistor.

$ΔUR=I2RΔt$ (XIII)

Here, $ΔUR$ is the energy delivered to the resistor, $R$ is the resistance, $I$ is the current in that branch.

Conclusion:

Substitute $0.846 A$ for $I$ , $8.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $8.00 Ω$ resistor.

$ΔUR=(0.846 A)2(8.00 Ω)(2.0 min×60 s1 min)=687 J$

Substitute $0.462 A$ for $I$ , $5.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $5.00 Ω$ resistor.

$ΔUR=(0.462 A)2(5.00 Ω)(2.0 min×60 s1 min)=128 J$

Substitute $0.462 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor.

$ΔUR=(0.462 A)2(1.00 Ω)(2.0 min×60 s1 min)=25.6 J$

Substitute $1.31 A$ for $I$ , $3.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $3.00 Ω$ resistor.

$ΔUR=(1.31 A)2(3.00 Ω)(2.0 min×60 s1 min)=616 J$

Substitute $1.31 A$ for $I$ , $1.00 Ω$ for $R$ , $2.00 min$ for $Δt$ in equation (XIII) to find energy delivered to $1.00 Ω$ resistor in the right hand branch.

$ΔUR=(1.31 A)2(1.00 Ω)(2.0 min×60 s1 min)=205 J$

Therefore, the energy delivered to $8.00 Ω$ resistor is $687 J_$ , $5.00 Ω$ resistor is $128 J_$ , $1.00 Ω$ resistor is $25.6 J_$ , $3.00 Ω$ resistor is $616 J_$ and $1.00 Ω$ resistor in the right hand branch is $205 J_$ .

Answer 29

(d)

Expert Solution

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Explanation of Solution

The chemical energy in the $12.0 V$ battery is transformed into internal energy in the resistors. The $4.00 V$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0 V$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The type of energy transformation occurs in the operation of the circuit.

Answer 34

Answer to Problem 47P

The chemical energy is transformed to the internal energy in the resistors.

Answer 35

Explanation of Solution

The chemical energy in the $12.0 V$ battery is transformed into internal energy in the resistors. The $4.00 V$ battery is being charged, so its chemical potential energy is increasing at the expense of some of the chemical potential energy in the $12.0 V$ battery.

Conclusion:

Therefore, the chemical energy is transformed to the internal energy in the resistors.

Answer 36

(e)

Expert Solution

To determine

The total amount of energy transformed into internal energy in the resistors.

Answer to Problem 47P

The total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

$ΔUT=ΔUB1+ΔUB2$ (XIV)

Here, $ΔUT$ is the total amount of energy transformed, $ΔUB1$ is the energy delivered to $4.00 V$ battery, $ΔUB2$ is the energy delivered to $12.0 V$ battery.

Conclusion:

Substitute $−222 J$ for $ΔUB1$ , $1.88 kJ$ for $ΔUB2$ in equation (XIV) to find $ΔUT$ .

$ΔUT=−222 J+(1.88 kJ×103J1 kJ)=1.66×103J=1.66 kJ$

Therefore, the total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

The total amount of energy transformed into internal energy in the resistors.

Answer 41

Answer to Problem 47P

The total amount of energy transformed into internal energy in the resistors is $1.66 kJ_$ .

Answer 42

Explanation of Solution

Write the expression for the total amount of energy transformed into internal energy in the resistors.

$ΔUT=ΔUB1+ΔUB2$ (XIV)

Here, $ΔUT$ is the total amount of energy transformed, $ΔUB1$ is the energy delivered to $4.00 V$ battery, $ΔUB2$ is the energy delivered to $12.0 V$ battery.