Chapter 25.2, Problem 16PP

To determine

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

Answer to Problem 16PP

Potential difference across the secondary circuit is $1.2\times {10}^2\text{V}$ and current in primary circuit $0.60\text{A}$ .

Explanation of Solution

Given:

  $V_p=7.2\text{kV}$

  $\begin{array}{l}{N}_s=125\\ N_p=7500\\ I_s=36\text{A}\end{array}$

Formula Used:

  $\frac{V_s}{V_p}=\frac{N_s}{N_p}$

Where,

  $V_p:$ Primary potential difference

  $V_s:$ Secondary potential difference

  $N_s:$ Number of turns of secondary coil

  $N_p:$ Number of turns of primary coil

  $V_p\times I_p=V_s\times I_s$

Where,

  $V_s:$ Secondary potential difference

  $V_p:$ Primary potential difference

  $I_s:$ Secondary current

  $I_p:$ Primary current

Calculation:

  $\begin{array}{c}{V}_p=7.2\text{kV}\\ =7.2\times {10}^3\text{V}\end{array}$

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

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

  $\frac{V_s}{V_p}=\frac{N_s}{N_p}$

Rearranging then above formula to get $V_s$ ,

  $V_s=\frac{V_p\times N_s}{N_p}$

Substituting the values in above formula

  $\begin{array}{l}{V}_s=\frac{\left(7.2\times {10}^3\text{V}\right)\left(125\right)}{7500}\\ =1.2\times {10}^2\text{V}\end{array}$

Current in primary circuit can be calculated by formula,

  $V_p\times I_p=V_s\times I_s$

Rearranging then above formula to get $I_p$ ,

  $I_p=\frac{V_s\times I_s}{V_p}$

Substituting the values in above formula

  $\begin{array}{c}{I}_p=\frac{\left(\text{1}{\text{.2×10}}^{\text{2}}\text{V}\right)\left(\text{36A}\right)}{\text{7}{\text{.2×10}}^{\text{3}}\text{V}}\\ =0.60\mathrm{A}\end{array}$

Conclusion:

Hence, thepotential difference across the secondary circuit is $1.2\times {10}^2\text{V}$ and current in primary circuit $0.60\text{A}\text{.}$