Engineering Fundamentals: An Introduction to Engineering
Engineering Fundamentals: An Introduction to Engineering
6th Edition
ISBN: 9780357391273
Author: Saeed Moaveni
Publisher: Cengage Learning US
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Chapter 12, Problem 19P
To determine

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

Expert Solution & Answer
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Answer to Problem 19P

The total resistance is 1.5Ω.

The current flows through the branch R1 is 2.0A.

The current flows through the branch R2 is 4.0A.

The current flows through the branch R3 is 2.0A.

Explanation of Solution

Given data:

The supply voltage is 12V.

The value of the resistor R1 is 6Ω.

The value of the resistor R2 is 3Ω.

The value of the resistor R3 is 6Ω.

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: An Introduction to Engineering, Chapter 12, Problem 19P

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

1Rtotal=16Ω+13Ω+16Ω=0.662Ω

Reduce the equation as,

Rtotal=10.662ΩRtotal=1.5Ω

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

V=VR1=VR2=VR3=12V

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

I1=12V6Ω=2A

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

I2=12V3Ω=4A

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

I3=12V6Ω=2A

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

Itotal=2Ω+4Ω+2Ω=8Ω

Conclusion:

Hence,

The total resistance is 1.5Ω.

The current flows through the branch R1 is 2.0A.

The current flows through the branch R2 is 4.0A.

The current flows through the branch R3 is 2.0A.

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