Chapter 25, Problem 57A

(a)

To determine

To calculate:

Number of turns that the Secondary coil has.

Answer to Problem 57A

The Secondary coil has 71 number of turns.

Explanation of Solution

Given:

Number of turns in primary coil = $475$

Potential difference = $18.$ V

Formula used:

Where,

   Effective potential difference of secondary coil

   No of turns of primary coil

   Effective potential difference of primary coil

   No of turns of secondary coil

Calculation:

No of turns that the secondary coil have can be calculated by the formula,

  $\begin{array}{l}\frac{V_s}{V_p}=\frac{N_s}{N_p}\\ N_s=\frac{V_s\times N_p}{V_p}\end{array}$

Substituting the values in the above equation,

  $\begin{array}{l}=\frac{(18.0)(475)}{120}\\ =71\end{array}$

Conclusion:

Hencethe secondary coil has 71numbers of turns.

(b)

To determine

To calculate:

Current in the primary circuit

Answer to Problem 57A

Current in the primary circuit is $18.75$ mA.

Explanation of Solution

Given:

  $V_s=18.0$ V= voltage in secondary coil.

  $I_s=125$ mA= current inn secondary coil.

  $V_p=120$ V= voltage in primary col.

Formula used:

  $V_p\times I_p=V_s\times I_s$

Where,

  $V_s=18.0$ V= Effective potential difference of secondary coil

  $I_s=125$ mA= Current of primary coil

  $V_p=120$ V= Effective potential difference of primary coil

  $I_p$ = Current of secondary coil

Calculation:

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

  $V_p\times I_p=V_s\times I_s$

Rearranging the equation,

  $I_p=\frac{V_s\times I_s}{V_p}$

Substituting the values in above formula,

  $\begin{array}{l}=\frac{(18.0)(125)}{120}\\ =18.75\end{array}$

  = $18.75$ mA

Conclusion:

Hencecurrent in the primary circuit is $18.75$ mA