Chapter 25, Problem 60A

(a)

To determine

To calculate:

Potential difference applied across the secondary circuit.

Answer to Problem 60A

Potential difference applied across the secondary circuit is $1.8$ kV.

Explanation of Solution

Given:

  $V_p$= $120$ V

  $N_s$ = $1200$

  $N_p$ = $80$

Formula used:

  $\frac{N_p}{N_s}=\frac{V_p}{V_s}$

where,

  $V_s$ = effective potential difference of secondary coil

  $N_p$ = No of turns of primary coil

  $V_p$ = effective potential difference of primary coil

  $N_s$ = No of turns of secondary coil

Calculation:

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

  $\frac{N_p}{N_s}=\frac{V_p}{V_s}$

Rearranging the above formula,

  $V_s=\frac{V_p\times N_s}{N_p}$

Substituting the values in above formula,

  $=\frac{(120)(1200)}{240}$

  $=1800$

  $=1.8$ kV

Conclusion:

Hencepotential difference applied across the secondary circuit is $1.8$ kV.

(b)

To determine

To calculate:

Current in the primary circuit

Answer to Problem 60A

Current in the primary circuit is $3.0\times 10$ A

Explanation of Solution

Given:

  $V_p=120$ V

  $V_s=1.8$ kV

  $I_s=2.0$ A

Formula used:

  $V_p{I}_p=V_s{I}_s$

where,

  $V_s$ = effective potential difference of secondary coil

  $N_p$ = No of turns of primary coil

  $V_p$ = effective potential difference of primary coil

  $N_s$ = No of turns of secondary coil

Calculation:

Current in the primary circuit can be calculated by the formula,

  $V_p{I}_p=V_s{I}_s$

Rearranging the above formula,

  $I_p=\frac{V_s{I}_s}{V_p}$

Substituting the values in above formula,

  $=\frac{(1.8\times {10}^3)(2.0)}{120}$

  $=3.0\times 10$ A

Conclusion:

Hencecurrent in the primary circuit is $=3.0\times 10$ A.

(c)

To determine

To calculate:

Power input and power output of transformer.

Answer to Problem 60A

Power input and power output of transformer are $3.6$ kWand $3.6$ kW.

Explanation of Solution

Given:

  $V_p=120$ V

  $V_s=1.8$ kV

  $I_s=2.0$ A

  $I_p=3.0\times 10$ A

Formula used:

  $\begin{array}{l}{V}_s{I}_S\\ V_p{I}_p\end{array}$

where,

  $V_s$ = effective potential difference of secondary coil

  $N_p$ = No of turns of primary coil

  $V_p$ = effective potential difference of primary coil

  $N_s$ = No of turns of secondary coil

Calculation:

Power input of transformer can be calculated by the formula,

  $V_p{I}_p$

Substituting the values in above formula,

  $\begin{array}{l}=(120)(30)\\ =3600\end{array}$

  $=3.6$ kW

Power output of transformer can be calculated by the formula,

  $V_s{I}_S$

Substituting the values in above formula,

  $\begin{array}{l}=(1800)(2.0)\\ =3600\\ =3.6\end{array}$

  $=3.6$ kW

Conclusion:

Hencepower input of transformer is and power output of transformer is