Chapter 25, Problem 74A

(a)

To determine

To calculate:

  $EMF$ of secondary circuit.

Answer to Problem 74A

  $EMF$ of secondary circuit is $3.6\times {10}^3$ V.

Explanation of Solution

Given:

  $V_p=120$ V

  $N_s=15000$

  $N_p=500$

Formula used:

  $\frac{V_p}{V_s}=\frac{N_p}{N_s}$

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:

  $EMF$ of secondary circuit can be calculated by the formula,

  $\frac{V_p}{V_s}=\frac{N_p}{N_s}$

Rearranging the above formula,

  $V_s=\frac{V_p\times N_s}{N_p}$

Substitutitng the values in above formula,

  $=\frac{(120)(15000)}{500}$

  $3.6\times {10}^3$ V

Conclusion:

Hence $EMF$ of secondary circuit is $3.6\times {10}^3$ V.

(b)

To determine

To calculate:

Current in the primary circuit.

Answer to Problem 74A

Current in the primary circuit is $9.0\times 10$ A.

Explanation of Solution

Given:

  $V_s=3600$ V

  $I_s=3.0$ A

  $V_p=120$ V

Formula used:

  $V_p{I}_p=V_s{I}_s$

where,

  $V_s$ = effective potential difference of secondary coil

  $I_s$ = current of primary coil

  $V_p$ = effective potential difference of primary coil

  $I_p$ = current 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,

  $=\frac{(3600)(3.0)}{120}$

  = $9.0\times 10$ A

Conclusion:

Hencecurrent in the primary circuit is $9.0\times 10$ A.

(c)

To determine

To calculate:

Power drawn by the primary circuit and secondary circuit.

Answer to Problem 74A

Power drawn by the primary circuit and secondary circuit is $1.1\times {10}^4$ W.

Explanation of Solution

Given:

  $V_s=3600$ V

  $I_s=3.0$ A

  $V_p=120$ V

  $I_p=9.0\times 10$ A

Formula used:

  $V_p{I}_p=V_s{I}_s$

where,

  $V_s$ = effective potential difference of secondary coil

  $I_s$ = current of primary coil

  $V_p$ = effective potential difference of primary coil

  $I_p$ = current of secondary coil

Calculation:

Power drawn by the primary circuit can be calculated by the formula,

= $V_p{I}_p$

Substituting the values in above formula,

  $=(120)(9.0\times 10)$

  = $=1.1\times {10}^4$ W

Power drawn by the secondary circuit can be calculated by the formula,

= $V_s{I}_s$

Substituting the values in above formula,

  $=(3600)(3.0)$

  $=1.1\times {10}^4$ W

Conclusion:

Hencepower drawn by the primary circuit and secondary circuit is $1.1\times {10}^4$ W.