Chapter 17, Problem 1R

What is the total current draw when all six fluorescent luminaires are turned on?

To determine

Find the total current drawn when all the six fluorescent luminaires are turned on.

Answer to Problem 1R

The total current $\left(I\right)$ that is drawn when all the six fluorescent luminaires are turned on is $9.2\text{A}$.

Explanation of Solution

Discussion:

The recreation room consists of six lay in fluorescent luminaires. One fluorescent luminaire consists of four $32\text{W}$ lamp.

$\begin{array}{c}\text{No}\text{.of watts for one luminaire}=4\times 32\text{W}\\ =128\text{W}\end{array}$

$\begin{array}{c}\text{No}\text{.of watts for six luminaire}=6\times 128\text{W}\\ =768\text{W}\end{array}$

Generally, the power factor of fluorescent luminaire lies between $0.5\text{and}\text{0}\text{.9}$. Let us assume the power factor $\left(\cos \varphi \right)$ as $0.762$.

$\text{No}\text{.of VA for six luminaire}=\frac{\text{No}\text{.of watts for six luminaire}}{\text{cos}\varphi }$

Substitute $768\text{W}$ for $\text{No}\text{.of watts for six luminaire}$ and $0.762$ for $\cos \varphi$ in the above equation to find $\text{No}\text{.of VA for six luminaire}$.

$\begin{array}{c}\text{No}\text{.of VA for six luminaire}=\frac{768\text{W}}{0.762}\\ =1008\text{VA}\left\{\because 1\text{W}=\text{1}\text{V}\cdot \text{1}\text{A}\right\}\end{array}$

The current drawn by the six fluorescent luminaires is calculated as follows:

$I=\frac{VA\text{for six luminaires}}{\text{single phase voltage}}$ (1)

Substitute $1008\text{VA}$ for $VA\text{for six luminaires}$ and $120\text{V}$ for $\text{single phase voltage}$ in equation (1) to find $I$.

$\begin{array}{c}I=\frac{1008\text{VA}}{\text{120}\text{V}}\\ =8.4\text{A}\end{array}$

One fluorescent luminaire is controlled by the 3-way switch. It also controls the stairway luminaire of $100\text{W}$.

When the staircase luminaire is turned on, the value of $\text{VA}$ becomes as follows:

$\begin{array}{c}\text{Total VA for six luminaires}=1008\text{VA}+100\text{W}\\ =1008\text{VA}+100\text{VA}\left\{\because 1\text{W}=1\text{V}\cdot \text{1A}\right\}\\ =1108\text{VA}\end{array}$

Substitute $1008\text{VA}$ for $\text{Total VA for six luminaires}$, and $120\text{V}$ for $\text{single phase voltage}$ to find $I$.

$\begin{array}{c}I=\frac{1108\text{VA}}{\text{120}\text{V}}\\ =9.2\text{A}\end{array}$

Conclusion:

Thus, the total current $\left(I\right)$ that is drawn when all the six fluorescent luminaires are turned on is $9.2\text{A}$.