Chapter 25, Problem 86A

(a)

To determine

To Calculate:Value of input power to the transformer.

Answer to Problem 86A

  $1.0\times {10}^2\text{kW}$

Explanation of Solution

Given:

Power consumed by load, $P_{out}=98\text{kW}$

Efficiency of transformer, $eff=98\%=0.98$

Formula used:

  $P_{in}=\frac{P_{out}}{eff}$

  $P_{in}:$ Input power

  $P_{out}:$ Output power

  $eff:$ Efficiency

Calculation:

Value of input power to the transformer can be calculated as follows:

  $P_{in}=\frac{P_{out}}{eff}$

Substitute the values in above formula:

  $\begin{array}{c}{P}_{in}=\frac{\text{98}\text{kW}}{\text{0}\text{.98}}\\ =1.0\times {10}^{\text{2}}\text{kW}\end{array}$

Conclusion:

Hencevalue of input power to the transformer is $1.0\times {10}^2\text{kW}$ .

(b)

To determine

To Calculate:

Maximum primary current when transformer delivers its rated power.

Answer to Problem 86A

Maximum primary current when transformer delivers its rated power is $200A$ .

Explanation of Solution

Given:

Input power, $P_{in}=1.0\times {10}^2\text{kW =100 kVA}$

Primary potential difference, $V_p=60\text{0}\text{V}$

Formula used:

  $I_p=\frac{P_{in}}{V_p}$

Where,

  $I_p:$ Maximum primary current

  $P_{in}:$ Input power

  $V_p:$ Primary potential difference

Calculation:

Maximum primary current when transformer delivers its rated power can be calculated as follows:

  $I_p=\frac{P_{in}}{V_p}$

Substituting the values in above formula,

  $\begin{array}{c}{I}_p=\frac{\text{100}\text{kVA}}{\text{600V}}\\ \text{=166}\text{.67A}\end{array}$

Conclusion:

Hence, maximum primary current when transformer delivers its rated power is $\text{166}\text{.67A}$ .