Chapter 25.2, Problem 17PP

To determine

To Calculate:Potential difference across the secondary circuit and current in the primary circuit

Answer to Problem 17PP

Potential difference across the secondary circuit is $1.80\times {10}^4\text{V}$ current in the primary circuit $1.6\times {10}^{-1}\text{A}$ .

Explanation of Solution

Given:

  $\begin{array}{c}{N}_s=90,000\\ N_p=300\\ V_p=60.0\text{V}\\ I_s=0.5\text{0}\text{A}\\ eff=95\%\\ =0.95\end{array}$

Formula Used:

  $\frac{V_s}{N_s}=\frac{V_p}{N_p}$

Where,

  $V_s:$ Secondary potential difference

  $V_p:$ Primary potential difference

  $N_s:$ No of turns of secondary coil

  $N_p:$ No of turns of primary coil

  $P_s=V_s\times I_s$

Where,

  $P_s:$ Secondary power

  $V_s:$ Secondary potential difference

  $I_s:$ Secondary current

  $P_p=\frac{P_s}{eff}$

Where,

  $P_p:$ Primary power

  $P_s:$ Secondary power

  $eff:$ Efficiency

  $I_p=\frac{P_p}{V_p}$

Where,

  $I_p:$ Primary current

  $P_p:$ Primary power

  $V_p:$ Primary potential difference

Calculation:

Potential difference across the secondary circuit can be calculated by formula,

  $\frac{V_s}{N_s}=\frac{V_p}{N_p}$

Rearranging the above formula to get $V_s$ ,

  $V_s=\frac{V_p\times N_s}{N_p}$

Substituting the values in above formula,

  $\begin{array}{c}{V}_s=\frac{\left(\text{60}\text{.0}\text{V}\right)\left(90,000\right)}{300}\\ =1.80\times {10}^4\text{V}\end{array}$

For secondary power,

  $P_s=V_s\times I_s$

Substituting the values in above formula,

  $\begin{array}{c}{P}_s=\left(\text{1}{\text{.80×10}}^{\text{4}}\text{V}\right)\left(\text{0}\text{.50A}\right)\\ =\text{9000}\text{W}\\ \text{=9}\text{.0}\text{kW}\end{array}$

As $1\text{kW=}1\times {10}^3\text{kW}$

For primary power,

  $P_p=\frac{P_s}{eff}$

Substituting the values in above formula,

  $\begin{array}{c}{P}_p=\frac{\text{9}\text{.0}\text{kW}}{\text{0}\text{.95}}\\ \text{=9}\text{.47}\text{kW}\end{array}$

For primary current,

  $I_p=\frac{P_p}{V_p}$

Substituting the values in above formula,

  $\begin{array}{c}{I}_p=\frac{\text{9}\text{.47}\text{kW}}{\text{60}\text{.0}\text{V}}\\ =1.6\times {10}^2\text{A}\end{array}$

Conclusion:

Hencepotential difference across the secondary circuit is $1.80\times {10}^4\text{V}$ current in the primary circuit is $1.6\times {10}^2\text{A}$ .