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.

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 $5\text{amperes}$.

Explanation of Solution

Given data:

The values of current and voltage at Load A are $10\text{amperes}$ and $120\text{volts}$.

The values of current and voltage at Load B are $5\text{amperes}$ and $120\text{volts}$.

The total voltage of the system is $240\text{volts}$.

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:

${\text{I}}_{\text{N}}={\text{I}}_{\text{A}}-{\text{I}}_{\text{B}}$ (1)

Substitute $10\text{amperes}$ for ${\text{I}}_{\text{A}}$, and $5\text{amperes}$ for ${\text{I}}_{\text{B}}$ in equation (1) to find ${\text{I}}_{\text{N}}$.

$\begin{array}{c}{\text{I}}_{\text{N}}=\left(10\text{amperes}\right)-\left(5\text{amperes}\right)\\ =5\text{amperes}\end{array}$

Conclusion:

Thus, the amount of current that flows in the neutral conductor is $5\text{amperes}$.

To determine

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

Answer to Problem 4R

The voltage of Load A is $80\text{volts}$ and the voltage of Load B is $160\text{volts}$.

Explanation of Solution

Formula used:

Write an expression to calculate the resistance.

$\text{R}=\frac{\text{E}}{\text{I}}$ . (2)

Here,

$\text{E}$ is the value of voltage, and

$\text{I}$ is the value of current.

Calculation:

Substitute $10\text{amperes}$ for ${\text{I}}_{\text{A}}$, and $120\text{volts}$ for ${\text{E}}_{\text{A}}$ in equation (2) to find ${\text{R}}_{\text{A}}$.

$\begin{array}{c}{\text{R}}_{\text{A}}=\frac{120\text{volts}}{10\text{amperes}}\\ =12\text{ohms}\end{array}$

Substitute $5\text{amperes}$ for ${\text{I}}_{\text{B}}$, and $120\text{volts}$ for ${\text{E}}_{\text{B}}$ in equation (2) to find ${\text{R}}_{\text{B}}$.

$\begin{array}{c}{\text{R}}_{\text{B}}=\frac{120\text{volts}}{5\text{amperes}}\\ =24\text{ohms}\end{array}$

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

${\text{R}}_{\text{T}}={\text{R}}_{\text{A}}+{\text{R}}_{\text{B}}$ (3)

Substitute $12\text{ohms}$ for ${\text{R}}_{\text{A}}$, and $24\text{ohms}$ for ${\text{R}}_{\text{B}}$ in equation (3) to find ${\text{R}}_{\text{T}}$.

$\begin{array}{c}{\text{R}}_{\text{T}}=12\text{ohms}+24\text{ohms}\\ =36\text{ohms}\end{array}$

Rearrange equation (2) to find $\text{I}$.

$\text{I}=\frac{\text{E}}{\text{R}}$ (4)

Substitute $240\text{volts}$ for $\text{E}$ and $36\text{ohms}$ for $\text{R}$ in equation (4) to find $\text{I}$.

$\begin{array}{c}\text{I}=\frac{240\text{volts}}{36\text{ohms}}\\ =6.66\text{amperes}\end{array}$

Rearrange equation (4) to find $\text{E}$.

$\text{E}=\text{IR}$ . (5)

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

Substitute $6.66\text{amperes}$ for $\text{I}$, and $12\text{ohms}$ for ${\text{R}}_{\text{A}}$ in equation (5) to find ${\text{E}}_{\text{A}}$.

$\begin{array}{c}{\text{E}}_{\text{A}}=\left(6.66\text{amperes}\right)\left(12\text{ohms}\right)\\ =80\text{volts}\end{array}$

Substitute $6.66\text{amperes}$ for $\text{I}$, and $24\text{ohms}$ for ${\text{R}}_{\text{B}}$ in equation (5) to find ${\text{E}}_{\text{B}}$.

$\begin{array}{c}{\text{E}}_{\text{B}}=\left(6.66\text{amperes}\right)\left(24\text{ohms}\right)\\ =160\text{volts}\end{array}$

Conclusion:

Thus, the voltage of Load A is $80\text{volts}$, and the voltage of Load B is $160\text{volts}$.