Chapter 24, Problem 1P

A balanced V load having a 10 $\Omega$ resistance in each leg is connected to a three-phase, four-wire. Y-connected generator having a line voltage of 208 V. Calculate the magnitude of

a. the phase voltage of the generator.

b. the phase voltage of the load.

c. the phase current of the load.

d. the line current.

To determine

(a)

Phase voltage of generator.

Answer to Problem 1P

The phase voltage of generator is $120.09\text{V}$.

Explanation of Solution

Given:

Resistance in each leg of $\text{Y}$ -connected load is $10\Omega$.

Line voltage of generator is $208\text{V}$.

Concept used:

Write the expression for phase voltage of the generator.

  $v_p=\frac{v_L}{\sqrt{3}}$ ...... (1)

Here, $v_p$ is phase voltage of generator and $v_L$ is line voltage of the generator.

Calculation:

Substitute $208\text{V}$ for $v_L$ in equation (1).

  $\begin{array}{c}{v}_p=\frac{208\text{V}}{\sqrt{3}}\\ =120.09\text{V}\end{array}$

Conclusion:

Thus, phase voltage of generator is $120.09\text{V}$.

To determine

(b)

Phase voltage of load.

Answer to Problem 1P

The value of Phase voltage of load is $120.09\text{V}$.

Explanation of Solution

Calculation:

As each phase of $\text{Y}$ -connected load is in parallel to each phase of the generator respectively so voltage across each phase of load is same as voltage across each phase of generator.

So value of Phase voltage of load is $120.09\text{V}$.

Conclusion:

Thus, value of Phase voltage of load is $120.09\text{V}$.

To determine

(c)

Phase current of load.

Answer to Problem 1P

The phase current of load is $12.01\text{A}$.

Explanation of Solution

Concept used:

Write the expression for phase current of load.

  $i_p=\frac{v_p}{R}$ ...... (2)

Here, $R$ is Resistance in each leg of $\text{Y}$ -connected load and $i_p$ is phase current of load.

Calculation:

Substitute $120.09\text{V}$ for $v_p$ and $10\Omega$ for $R$ in equation (2).

  $\begin{array}{c}{i}_p=\frac{120.09\text{V}}{10\Omega }\\ =12.01\text{A}\end{array}$

Conclusion:

Thus, phase current of load is $12.01\text{A}$.

To determine

(d)

Value of line current.

Answer to Problem 1P

The value of line current is $12.01\text{A}$.

Explanation of Solution

Calculation:

Value of line current is equal to value of phase current in $\text{Y}$ -connected load.

As load is $\text{Y}$ -connected and value of phase current is $12.01\text{A}$ so value of line current is $12.01\text{A}$.

Conclusion:

Thus, value of line current is $12.01\text{A}$.