EBK ELECTRICAL WIRING RESIDENTIAL
EBK ELECTRICAL WIRING RESIDENTIAL
19th Edition
ISBN: 9781337516549
Author: Simmons
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 17, Problem 4R

In the diagram, Load A is rated at 10 amperes, 120 volts. Load B is rated at 5 amperes, 120 volts.

Chapter 17, Problem 4R, In the diagram, Load A is rated at 10 amperes, 120 volts. Load B is rated at 5 amperes, 120 volts.

  1. a. When connected to the 3-wire branch circuit as indicated, how much current will flow in the neutral conductor? __________________
  2. b. If the neutral conductor should open, to what voltage would each load be subjected, assuming both loads were operating at the time the neutral conductor opened? Show all calculations.

a.

Expert Solution
Check Mark
To determine

Find the amount of current that flows in a neutral conductor when a circuit is connected to a 3-wire branch circuit.

Answer to Problem 4R

The amount of current that flows in the neutral conductor is 5amperes.

Explanation of Solution

Given data:

The values of current and voltage at Load A are 10amperes and 120volts.

The values of current and voltage at Load B are 5amperes and 120volts.

The total voltage of the system is 240volts.

Calculation:

From the given data, the values of current to Load A and Load B are not equal. Therefore, it is an unbalanced load.

For an unbalanced load, the current through the neutral conductor is calculated as follows:

IN=IAIB (1)

Substitute 10amperes for IA, and 5amperes for IB in equation (1) to find IN.

IN=(10amperes)(5amperes)=5amperes

Conclusion:

Thus, the amount of current that flows in the neutral conductor is 5amperes.

b.

Expert Solution
Check Mark
To determine

Find the voltage of each load when the neutral conductor is open.

Answer to Problem 4R

The voltage of Load A is 80volts and the voltage of Load B is 160volts.

Explanation of Solution

Formula used:

Write an expression to calculate the resistance.

R=EI . (2)

Here,

E is the value of voltage, and

I is the value of current.

Calculation:

Substitute 10amperes for IA, and 120volts for EA in equation (2) to find RA.

RA=120volts10amperes=12ohms

Substitute 5amperes for IB, and 120volts for EB in equation (2) to find RB.

RB=120volts5amperes=24ohms

When the neutral conductor is open, Load A and Load B are connected in series. Therefore, the total resistance is calculated as follows:

RT=RA+RB (3)

Substitute 12ohms for RA, and 24ohms for RB in equation (3) to find RT.

RT=12ohms+24ohms=36ohms

Rearrange equation (2) to find I.

I=ER (4)

Substitute 240volts for E and 36ohms for R in equation (4) to find I.

I=240volts36ohms=6.66amperes

Rearrange equation (4) to find E.

E=IR . (5)

For the series connected load, the current is same for both Load A and Load B.

Substitute 6.66amperes for I, and 12ohms for RA in equation (5) to find EA.

EA=(6.66amperes)(12ohms)=80volts

Substitute 6.66amperes for I, and 24ohms for RB in equation (5) to find EB.

EB=(6.66amperes)(24ohms)=160volts

Conclusion:

Thus, the voltage of Load A is 80volts, and the voltage of Load B is 160volts.

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EBK ELECTRICAL WIRING RESIDENTIAL
Electrical Engineering
ISBN:9781337516549
Author:Simmons
Publisher:CENGAGE LEARNING - CONSIGNMENT
Demos: Dielectric breakdown; Author: Caltech's Feynman Lecture Hall;https://www.youtube.com/watch?v=2YrHh1ikefI;License: Standard Youtube License