Chapter 21, Problem 46QAP

To determine

(a)

The ratio of number of turns in the primary and the secondary coil.

Answer to Problem 46QAP

The ratio of number of turns in secondary and the primary coil is $41$.

Explanation of Solution

Given info:

The primary voltage is $\text{120}{\text{V}}_{rms}$.

The secondary voltage is $\text{5000}{\text{V}}_{rms}$.

The power dissipated value is $75\text{W}$.

Formula used:

The formula to find the ratio of number of turns in secondary and the primary coil is,

  $\frac{N_s}{N_p}=\frac{V_s}{V_p}$   Equation-1

Here, $V_p$ is the primary voltage and $N_s$ is the number of turns in the secondary coil and $N_p$ is the number of turns in the primary coil.

Calculation:

Substitute the given values in equation-1 to find the ratio.

  $\begin{array}{c}\frac{N_s}{N_p}=\frac{5000{\text{V}}_{\text{rms}}{}_{}}{120{\text{V}}_{\text{rms}}}\\ =41\text{Equation-2}\end{array}$

Conclusion:

From equation -2 the ratio of number of turns in secondary and the primary coil is $41$.

To determine

(b)

The value of the rms current in the primary and the secondary coil.

Answer to Problem 46QAP

The value of the rms current in the secondary coil is $0.015\text{A}$ and the primary coil is $0.625\text{A}$.

Explanation of Solution

Given info:

The primary voltage is $\text{120}{\text{V}}_{rms}$.

The secondary voltage is $\text{5000}{\text{V}}_{rms}$.

The power dissipated value is $75\text{W}$.

Formula used:

The formula to find the rms current is,

  $I_{rms}=\frac{P}{V_{rms}}$   Equation-3

Here, $V_{rms}$ is the rms voltage and $P$ is the power dissipated.

Calculation:

Substitute the given values in equation-1 to find the rms current in the primary and secondary coil.

Primary coil rms current $I_{p(rms)}$ is,

  $\begin{array}{c}{I}_{p(rms)}=\frac{75W}{120{\text{V}}_{\text{rms}}}\\ =0.625\text{A}\text{Equation-4}\end{array}$

Secondary coil rms current $I_{s(rms)}$ is,

  $\begin{array}{c}{I}_{s(rms)}=\frac{75W}{5000{\text{V}}_{\text{rms}}}\\ =0.015\text{A}\text{Equation-5}\end{array}$

Conclusion:

From equation -4 and equation-5 the rms current in the secondary coil is $0.015\text{A}$ and the primary coil is $0.625\text{A}$.

To determine

(c)

The the effective resistance of the primary coil.

Answer to Problem 46QAP

The effective resistance of the primary coil is $192\Omega$.

Explanation of Solution

Given info:

The primary voltage is $\text{120}{\text{V}}_{rms}$.

The secondary voltage is $\text{5000}{\text{V}}_{rms}$.

The power dissipated value is $75\text{W}$.

Formula used:

The formula to find the effective resistance of the primary coil is,

  $R_{\text{p}}=\frac{V_{\text{p(rms)}}{}^2}{P}$   Equation-3

Here, $V_{\text{p(rms)}}$ is the rms voltage of the primary coil and $P$ is the power dissipated.

Calculation:

Substitute the given values in equation-1 to find the effective resistance.

Primary coil rms current $I_{p(rms)}$ is,

  $\begin{array}{c}{R}_{\text{p}}=\frac{{\left(120{\text{V}}_{\text{rms}}\right)}^2}{75W}\\ =192\Omega \text{Equation-6}\end{array}$

Conclusion:

From equation-6 effective resistance of the primary coil is $192\Omega$.