Chapter 37, Problem 58A

(a)

To determine

To calculate: the AC voltage available in secondary coil.

Answer to Problem 58A

The voltage in the secondary coil is $60V$ .

Explanation of Solution

Given:

Number of turns on primary coil $\left(N_1\right)=50$

Number of turns on secondary coil $\left(N_2\right)=250.$

Voltage on primary coil $\left(V_1\right)=12V$

Voltage in secondary coil $\left(V_2\right)=?$

Formula used:

  $\frac{V_1}{V_2}=\frac{N_1}{N_2}$.

Calculation:

The relation between voltages and number on turns on primary and secondary coil is,

  $\frac{V_1}{V_2}=\frac{N_1}{N_2}$

The equation becomes,

  $\begin{array}{l}{V}_2=V_1\frac{N_2}{N_1}\\ V_2=12\times \frac{250}{50}\\ =60volts\end{array}$

The voltage in secondary coil is $60V$

Conclusion:  The voltage in secondary coil is $60V$ calculated by using relation between number of turns and the voltage in the transformer.

(b)

To determine

To calculate: the current in the $10ohms$ resistor.

Answer to Problem 58A

The current in the $10ohms$ resistor is $6A.$

Explanation of Solution

Given:

Voltage in the secondary coil $\left(V_2\right)=60V$

Resistance in secondary circuit $\left(R\right)=10\Omega$

Formula used:

  $\mathrm{I}=\frac{V}{R}$

Calculation:

Ohm’s law states that in any electric circuit the current is directly proportional to voltage and inversely proportional to resistance.

Mathematically,

  $\mathrm{I}=\frac{V}{R}$ .

  $\begin{array}{l}I=\frac{60}{10}A\\ I=6A\end{array}$

The current in the $10ohms$ circuit is $6A.$

Conclusion:

The current in the circuit is $6A.$

(c)

To determine

To calculate: the power supply in the primary circuit.

Answer to Problem 58A

The power is $360W.$

Explanation of Solution

Given:

Voltage in the primary coil $\left(V_2\right)=60V$

Current in secondary circuit $\left(I_2\right)=6A$

Formula used:

Power into the primary coil is equal to the Power out from the secondary coil. And the formula of the Power is given by the relation between voltage and current. Measured in watts(W).

Mathematically,

  $\mathrm{P}=VIWatts$.

  $\mathrm{P}=(60V)\times (6A)W$

$\mathrm{P}=360W$.$P=360\text{ W}$

The power supplied to the primary circuit is $360W.$

Conclusion:

By the concept of the power out from the secondary coil to the primary coil, the power calculated as $360W.$