Chapter 25, Problem 85A

To determine

To Calculate:Effective primary current.

Answer to Problem 85A

  $4.\text{7A}$

Explanation of Solution

Given:

Secondary potential difference, $V_{s,peak}=60.0\text{V}$

Number of secondary turns, $N_S=10$

Number of primary turns, $N_P=100$

Power of load, $P=2.\text{0}\text{kW}$

  $2.\text{0}\text{kW}=2.0\times {10}^3\text{W}$

As,

  $\text{1}{\text{kW = 1×10}}^{\text{3}}\text{W}$

Formula Used:

  $V_{s,eff}=\frac{V_{s,peak}}{\sqrt{2}}$

Where,

  $V_{s,eff}:$ Secondary effective potential difference

  $V_{s,peak}:$ Secondary peak potential difference

  $I_{s,eff}=\frac{P}{V_{s,eff}}$

Where,

  $I_{s,eff}:$ Secondary effective current

  $P:$ Power

  $V_{s,eff}:$ Secondary peak potential difference

  $I_{p,eff}=\left(\frac{N_s}{N_p}\right)\times I_{s,eff}$

Where,

  $I_{p,eff}:$ Primary effective current

  $I_{s,eff}:$ Secondary effective current

  $N_s:$ No of turns for secondary coil

  $N_p:$ No of turns for primary coil

Calculation:

For secondary effective potential difference,

  $V_{s,eff}=\frac{V_{s,peak}}{\sqrt{2}}$

Substituting the values in above formula,

  $\begin{array}{c}{V}_{s,eff}=\frac{\text{60}\text{.0}\text{V}}{\sqrt{\text{2}}}\\ =\text{42}\text{.4}\text{V}\end{array}$

For secondary effective current,

  $I_{s,eff}=\frac{P}{V_{s,eff}}$

Substituting the values in above formula,

  $\begin{array}{c}{I}_{s,eff}\text{=}\frac{\text{2}{\text{.0×10}}^{\text{3}}\text{W}}{\text{42}\text{.4}\text{V}}\\ \text{=}\text{47}\text{A}\end{array}$

Effective primary current can be calculated using the formula:

  $I_{p,eff}=\left(\frac{N_s}{N_p}\right)\times I_{s,eff}$

Substituting the values in above formula,

  $\begin{array}{c}{I}_{p,eff}=\left(\frac{10}{100}\right)\times 47\text{A}\\ =\text{4}\text{.7A}\end{array}$

Conclusion:

Hence,effective primary current is $4.7\text{A}$ .