Find the total resistance and the current flow in each branch for the circuit.

Question

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Answer 1

Question

Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Expert Solution & Answer

Answer to Problem 19P

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Explanation of Solution

Given data:

The supply voltage is $12 V$ .

The value of the resistor $R1$ is $6 Ω$ .

The value of the resistor $R2$ is $3 Ω$ .

The value of the resistor $R3$ is $6 Ω$ .

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$1Rtotal=1R1+1R2+1R3$ (1)

Here,

$R1$ , $R2$ , $R3$ are the resistances.

Formula to calculate the voltage across the resistor $R1$ ,

$V=I1R1$

Here,

$I1$ is the current flow through the resistance $R1$ .

$R1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R1$ ,

$I1=VR1$ (2)

Formula to calculate the voltage across the resistor $R2$ ,

$V=I2R2$

Here,

$R2$ is the resistance.

$I2$ is the current flow through the resistance $R2$ .

Rearrange the equation for the current flow through the resistance $R2$ ,

$I2=VR2$ (3)

Formula to calculate the voltage across the resistor $R3$ ,

$V=I3R3$

Here,

$R3$ is the resistance.

$I3$ is the current flow through the resistance $R3$ .

Rearrange the equation for the current flow through the resistance $R3$ ,

$I3=VR3$ (4)

Formula to calculate the total current drawn by the circuit,

$Itotal=I1+I2+I3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $(R1,R3)$ are connected in a parallel arrangement with the resistor $R2$ .

ENGINEERING FUNDAMENTALS, Chapter 12, Problem 19P

Substitute $6 Ω$ for $R1$ , $3 Ω$ for $R2$ and $6 Ω$ for $R3$ in equation (1) to find $Rtotal$ ,

$1Rtotal=16 Ω+13 Ω+16 Ω=0.662 Ω$

Reduce the equation as,

$Rtotal=10.662 ΩRtotal=1.5 Ω$

The voltage drop across the each light bulb and the resistor is equal to the $12 V$ .

$V=VR1=VR2=VR3=12 V$

Substitute $12 V$ for V and $6 Ω$ for $R1$ in equation (2) to find $I1$ ,

$I1=12 V6 Ω=2 A$

Substitute $12 V$ for V and $3 Ω$ for $R2$ in equation (3) to find $I2$ ,

$I2=12 V3 Ω=4 A$

Substitute $12 V$ for V and $6 Ω$ for $R3$ in equation (4) to find $I3$ ,

$I3=12 V6 Ω=2 A$

Substitute $2 A$ for $I1$ , $4 A$ for $I2$ and $2 A$ for $I3$ in equation (5) to find $Itotal$ ,

$Itotal=2 Ω+4 Ω+2 Ω=8 Ω$

Conclusion:

Hence,

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

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Answer 2

Question

Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Expert Solution & Answer

Answer to Problem 19P

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Explanation of Solution

Given data:

The supply voltage is $12 V$ .

The value of the resistor $R1$ is $6 Ω$ .

The value of the resistor $R2$ is $3 Ω$ .

The value of the resistor $R3$ is $6 Ω$ .

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$1Rtotal=1R1+1R2+1R3$ (1)

Here,

$R1$ , $R2$ , $R3$ are the resistances.

Formula to calculate the voltage across the resistor $R1$ ,

$V=I1R1$

Here,

$I1$ is the current flow through the resistance $R1$ .

$R1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R1$ ,

$I1=VR1$ (2)

Formula to calculate the voltage across the resistor $R2$ ,

$V=I2R2$

Here,

$R2$ is the resistance.

$I2$ is the current flow through the resistance $R2$ .

Rearrange the equation for the current flow through the resistance $R2$ ,

$I2=VR2$ (3)

Formula to calculate the voltage across the resistor $R3$ ,

$V=I3R3$

Here,

$R3$ is the resistance.

$I3$ is the current flow through the resistance $R3$ .

Rearrange the equation for the current flow through the resistance $R3$ ,

$I3=VR3$ (4)

Formula to calculate the total current drawn by the circuit,

$Itotal=I1+I2+I3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $(R1,R3)$ are connected in a parallel arrangement with the resistor $R2$ .

ENGINEERING FUNDAMENTALS, Chapter 12, Problem 19P

Substitute $6 Ω$ for $R1$ , $3 Ω$ for $R2$ and $6 Ω$ for $R3$ in equation (1) to find $Rtotal$ ,

$1Rtotal=16 Ω+13 Ω+16 Ω=0.662 Ω$

Reduce the equation as,

$Rtotal=10.662 ΩRtotal=1.5 Ω$

The voltage drop across the each light bulb and the resistor is equal to the $12 V$ .

$V=VR1=VR2=VR3=12 V$

Substitute $12 V$ for V and $6 Ω$ for $R1$ in equation (2) to find $I1$ ,

$I1=12 V6 Ω=2 A$

Substitute $12 V$ for V and $3 Ω$ for $R2$ in equation (3) to find $I2$ ,

$I2=12 V3 Ω=4 A$

Substitute $12 V$ for V and $6 Ω$ for $R3$ in equation (4) to find $I3$ ,

$I3=12 V6 Ω=2 A$

Substitute $2 A$ for $I1$ , $4 A$ for $I2$ and $2 A$ for $I3$ in equation (5) to find $Itotal$ ,

$Itotal=2 Ω+4 Ω+2 Ω=8 Ω$

Conclusion:

Hence,

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

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Answer 3

Question

Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Expert Solution & Answer

Answer to Problem 19P

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Explanation of Solution

Given data:

The supply voltage is $12 V$ .

The value of the resistor $R1$ is $6 Ω$ .

The value of the resistor $R2$ is $3 Ω$ .

The value of the resistor $R3$ is $6 Ω$ .

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$1Rtotal=1R1+1R2+1R3$ (1)

Here,

$R1$ , $R2$ , $R3$ are the resistances.

Formula to calculate the voltage across the resistor $R1$ ,

$V=I1R1$

Here,

$I1$ is the current flow through the resistance $R1$ .

$R1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R1$ ,

$I1=VR1$ (2)

Formula to calculate the voltage across the resistor $R2$ ,

$V=I2R2$

Here,

$R2$ is the resistance.

$I2$ is the current flow through the resistance $R2$ .

Rearrange the equation for the current flow through the resistance $R2$ ,

$I2=VR2$ (3)

Formula to calculate the voltage across the resistor $R3$ ,

$V=I3R3$

Here,

$R3$ is the resistance.

$I3$ is the current flow through the resistance $R3$ .

Rearrange the equation for the current flow through the resistance $R3$ ,

$I3=VR3$ (4)

Formula to calculate the total current drawn by the circuit,

$Itotal=I1+I2+I3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $(R1,R3)$ are connected in a parallel arrangement with the resistor $R2$ .

ENGINEERING FUNDAMENTALS, Chapter 12, Problem 19P

Substitute $6 Ω$ for $R1$ , $3 Ω$ for $R2$ and $6 Ω$ for $R3$ in equation (1) to find $Rtotal$ ,

$1Rtotal=16 Ω+13 Ω+16 Ω=0.662 Ω$

Reduce the equation as,

$Rtotal=10.662 ΩRtotal=1.5 Ω$

The voltage drop across the each light bulb and the resistor is equal to the $12 V$ .

$V=VR1=VR2=VR3=12 V$

Substitute $12 V$ for V and $6 Ω$ for $R1$ in equation (2) to find $I1$ ,

$I1=12 V6 Ω=2 A$

Substitute $12 V$ for V and $3 Ω$ for $R2$ in equation (3) to find $I2$ ,

$I2=12 V3 Ω=4 A$

Substitute $12 V$ for V and $6 Ω$ for $R3$ in equation (4) to find $I3$ ,

$I3=12 V6 Ω=2 A$

Substitute $2 A$ for $I1$ , $4 A$ for $I2$ and $2 A$ for $I3$ in equation (5) to find $Itotal$ ,

$Itotal=2 Ω+4 Ω+2 Ω=8 Ω$

Conclusion:

Hence,

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

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Answer 4

Question

Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Expert Solution & Answer

Answer to Problem 19P

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Explanation of Solution

Given data:

The supply voltage is $12 V$ .

The value of the resistor $R1$ is $6 Ω$ .

The value of the resistor $R2$ is $3 Ω$ .

The value of the resistor $R3$ is $6 Ω$ .

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$1Rtotal=1R1+1R2+1R3$ (1)

Here,

$R1$ , $R2$ , $R3$ are the resistances.

Formula to calculate the voltage across the resistor $R1$ ,

$V=I1R1$

Here,

$I1$ is the current flow through the resistance $R1$ .

$R1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R1$ ,

$I1=VR1$ (2)

Formula to calculate the voltage across the resistor $R2$ ,

$V=I2R2$

Here,

$R2$ is the resistance.

$I2$ is the current flow through the resistance $R2$ .

Rearrange the equation for the current flow through the resistance $R2$ ,

$I2=VR2$ (3)

Formula to calculate the voltage across the resistor $R3$ ,

$V=I3R3$

Here,

$R3$ is the resistance.

$I3$ is the current flow through the resistance $R3$ .

Rearrange the equation for the current flow through the resistance $R3$ ,

$I3=VR3$ (4)

Formula to calculate the total current drawn by the circuit,

$Itotal=I1+I2+I3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $(R1,R3)$ are connected in a parallel arrangement with the resistor $R2$ .

ENGINEERING FUNDAMENTALS, Chapter 12, Problem 19P

Substitute $6 Ω$ for $R1$ , $3 Ω$ for $R2$ and $6 Ω$ for $R3$ in equation (1) to find $Rtotal$ ,

$1Rtotal=16 Ω+13 Ω+16 Ω=0.662 Ω$

Reduce the equation as,

$Rtotal=10.662 ΩRtotal=1.5 Ω$

The voltage drop across the each light bulb and the resistor is equal to the $12 V$ .

$V=VR1=VR2=VR3=12 V$

Substitute $12 V$ for V and $6 Ω$ for $R1$ in equation (2) to find $I1$ ,

$I1=12 V6 Ω=2 A$

Substitute $12 V$ for V and $3 Ω$ for $R2$ in equation (3) to find $I2$ ,

$I2=12 V3 Ω=4 A$

Substitute $12 V$ for V and $6 Ω$ for $R3$ in equation (4) to find $I3$ ,

$I3=12 V6 Ω=2 A$

Substitute $2 A$ for $I1$ , $4 A$ for $I2$ and $2 A$ for $I3$ in equation (5) to find $Itotal$ ,

$Itotal=2 Ω+4 Ω+2 Ω=8 Ω$

Conclusion:

Hence,

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

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Answer 5

Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Expert Solution & Answer

Answer to Problem 19P

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Explanation of Solution

Given data:

The supply voltage is $12 V$ .

The value of the resistor $R1$ is $6 Ω$ .

The value of the resistor $R2$ is $3 Ω$ .

The value of the resistor $R3$ is $6 Ω$ .

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$1Rtotal=1R1+1R2+1R3$ (1)

Here,

$R1$ , $R2$ , $R3$ are the resistances.

Formula to calculate the voltage across the resistor $R1$ ,

$V=I1R1$

Here,

$I1$ is the current flow through the resistance $R1$ .

$R1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R1$ ,

$I1=VR1$ (2)

Formula to calculate the voltage across the resistor $R2$ ,

$V=I2R2$

Here,

$R2$ is the resistance.

$I2$ is the current flow through the resistance $R2$ .

Rearrange the equation for the current flow through the resistance $R2$ ,

$I2=VR2$ (3)

Formula to calculate the voltage across the resistor $R3$ ,

$V=I3R3$

Here,

$R3$ is the resistance.

$I3$ is the current flow through the resistance $R3$ .

Rearrange the equation for the current flow through the resistance $R3$ ,

$I3=VR3$ (4)

Formula to calculate the total current drawn by the circuit,

$Itotal=I1+I2+I3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $(R1,R3)$ are connected in a parallel arrangement with the resistor $R2$ .

ENGINEERING FUNDAMENTALS, Chapter 12, Problem 19P

Substitute $6 Ω$ for $R1$ , $3 Ω$ for $R2$ and $6 Ω$ for $R3$ in equation (1) to find $Rtotal$ ,

$1Rtotal=16 Ω+13 Ω+16 Ω=0.662 Ω$

Reduce the equation as,

$Rtotal=10.662 ΩRtotal=1.5 Ω$

The voltage drop across the each light bulb and the resistor is equal to the $12 V$ .

$V=VR1=VR2=VR3=12 V$

Substitute $12 V$ for V and $6 Ω$ for $R1$ in equation (2) to find $I1$ ,

$I1=12 V6 Ω=2 A$

Substitute $12 V$ for V and $3 Ω$ for $R2$ in equation (3) to find $I2$ ,

$I2=12 V3 Ω=4 A$

Substitute $12 V$ for V and $6 Ω$ for $R3$ in equation (4) to find $I3$ ,

$I3=12 V6 Ω=2 A$

Substitute $2 A$ for $I1$ , $4 A$ for $I2$ and $2 A$ for $I3$ in equation (5) to find $Itotal$ ,

$Itotal=2 Ω+4 Ω+2 Ω=8 Ω$

Conclusion:

Hence,

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Answer 6

Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Expert Solution & Answer

Answer to Problem 19P

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Explanation of Solution

Given data:

The supply voltage is $12 V$ .

The value of the resistor $R1$ is $6 Ω$ .

The value of the resistor $R2$ is $3 Ω$ .

The value of the resistor $R3$ is $6 Ω$ .

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$1Rtotal=1R1+1R2+1R3$ (1)

Here,

$R1$ , $R2$ , $R3$ are the resistances.

Formula to calculate the voltage across the resistor $R1$ ,

$V=I1R1$

Here,

$I1$ is the current flow through the resistance $R1$ .

$R1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R1$ ,

$I1=VR1$ (2)

Formula to calculate the voltage across the resistor $R2$ ,

$V=I2R2$

Here,

$R2$ is the resistance.

$I2$ is the current flow through the resistance $R2$ .

Rearrange the equation for the current flow through the resistance $R2$ ,

$I2=VR2$ (3)

Formula to calculate the voltage across the resistor $R3$ ,

$V=I3R3$

Here,

$R3$ is the resistance.

$I3$ is the current flow through the resistance $R3$ .

Rearrange the equation for the current flow through the resistance $R3$ ,

$I3=VR3$ (4)

Formula to calculate the total current drawn by the circuit,

$Itotal=I1+I2+I3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $(R1,R3)$ are connected in a parallel arrangement with the resistor $R2$ .

ENGINEERING FUNDAMENTALS, Chapter 12, Problem 19P

Substitute $6 Ω$ for $R1$ , $3 Ω$ for $R2$ and $6 Ω$ for $R3$ in equation (1) to find $Rtotal$ ,

$1Rtotal=16 Ω+13 Ω+16 Ω=0.662 Ω$

Reduce the equation as,

$Rtotal=10.662 ΩRtotal=1.5 Ω$

The voltage drop across the each light bulb and the resistor is equal to the $12 V$ .

$V=VR1=VR2=VR3=12 V$

Substitute $12 V$ for V and $6 Ω$ for $R1$ in equation (2) to find $I1$ ,

$I1=12 V6 Ω=2 A$

Substitute $12 V$ for V and $3 Ω$ for $R2$ in equation (3) to find $I2$ ,

$I2=12 V3 Ω=4 A$

Substitute $12 V$ for V and $6 Ω$ for $R3$ in equation (4) to find $I3$ ,

$I3=12 V6 Ω=2 A$

Substitute $2 A$ for $I1$ , $4 A$ for $I2$ and $2 A$ for $I3$ in equation (5) to find $Itotal$ ,

$Itotal=2 Ω+4 Ω+2 Ω=8 Ω$

Conclusion:

Hence,

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Answer 7

Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Answer 8

Expert Solution & Answer

Answer to Problem 19P

The total resistance is $1.5 Ω$ .

The current flows through the branch $R1$ is $2.0 A$ .

The current flows through the branch $R2$ is $4.0 A$ .

The current flows through the branch $R3$ is $2.0 A$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

The supply voltage is $12 V$ .

The value of the resistor $R1$ is $6 Ω$ .

The value of the resistor $R2$ is $3 Ω$ .

The value of the resistor $R3$ is $6 Ω$ .

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$1Rtotal=1R1+1R2+1R3$ (1)

Here,

$R1$ , $R2$ , $R3$ are the resistances.

Formula to calculate the voltage across the resistor $R1$ ,

$V=I1R1$

Here,

$I1$ is the current flow through the resistance $R1$ .

$R1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R1$ ,

$I1=VR1$ (2)

Formula to calculate the voltage across the resistor $R2$ ,

$V=I2R2$

Here,

$R2$ is the resistance.

$I2$ is the current flow through the resistance $R2$ .

Rearrange the equation for the current flow through the resistance $R2$ ,

$I2=VR2$ (3)

Formula to calculate the voltage across the resistor $R3$ ,

$V=I3R3$

Here,

$R3$ is the resistance.

$I3$ is the current flow through the resistance $R3$ .

Rearrange the equation for the current flow through the resistance $R3$ ,

$I3=VR3$ (4)

Formula to calculate the total current drawn by the circuit,

$Itotal=I1+I2+I3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $(R1,R3)$ are connected in a parallel arrangement with the resistor $R2$ .