Chapter 23, Problem 1P

For the air-core transformer in Fig. 23.57:

1. Find the value of L_s, if the mutual inductance M is equal to 40 mH.
2. Find the induced voltages $e_p$
3. and $e_s$
4. if the flux linking the primary coil changes at the rate of 0.08 Wb/s.
5. Find the induced voltages $e_p\text{ and }{e}_s\text{ if the current }{i}_p$
6. changes at the rate of 0.3 A/ms.

To determine

(a)

Value of $L_s$ for the given mutual inductance.

Answer to Problem 1P

Value of $L_s$ is $50\text{mH}$.

Explanation of Solution

Given:

  

Mutual inductance, M = 40mH

Formula used:

Value of $L_s$ is calculated by

  $M=k\sqrt{L_p{L}_s}$

Calculation:

With the known values, value of $L_s$ is calculated as

  $M=k\sqrt{L_p{L}_s}$

We know that ${\text{L}}_p=50mH,M=40mH\&k=0.8$

  $\begin{array}{l}(40\times {10}^{-3})=0.8\sqrt{(50\times {10}^{-3})L_s}\\ $ (50\times {10}^{-3})L_s={(\frac{(40\times {10}^{-3})}{0.8})}^2$$ L_s={(\frac{(40\times {10}^{-3})}{0.8})}^2(\frac{1}{50\times {10}^{-3}})$=.05\\ L_s=50mH\end{array}$

Conclusion:

Thus, the value of $L_s$ is $50mH$

To determine

(b)

The value of the primary and secondary induced voltage at the rate of $0.08Wb/s$.

Answer to Problem 1P

The value of primary induced voltage is 1.6V

The value of secondary induced voltage is 5.12V

Explanation of Solution

Given:

  

  $\frac{d\varphi }{dt}=0.08Wb/s$

Formula used:

The primary induced voltage is calculated by

  $e_p=N_p\frac{d\varphi }{dt}$

The secondary induced voltage is calculated by

  $e_s=k{N}_s\frac{d\varphi }{dt}$

Calculation:

With the known values, the primary induced voltage is calculated as

  $e_p=N_p\frac{d\varphi }{dt}$

Substitute ${\text{N}}_p=20\text{and }\frac{d\varphi }{dt}=0.08\text{Wb/s}$

  $e_p=(20)(0.08\text{Wb/s})=1.6\text{V}$

The value of primary induced voltage is 1.6V

Secondary induced voltage,

  $e_s=k{N}_s\frac{d\varphi }{dt}$

Substitute ${\text{N}}_s=80,k=0.8\text{and}\frac{d\varphi }{dt}=0.08\text{Wb/s}$

  $e_s=(0.8)(80)(0.08Wb/s)=5.12V$

The value of secondary induced voltage is 5.12V

Conclusion:

Thus, the value of primary induced voltage is 1.6V and the value of secondary induced voltage is 5.12V

To determine

(c)

The value of the primary and secondary induced voltage at the rate of $0.3A/ms$

Answer to Problem 1P

Value of primary induced voltage is 15V

Value of secondary induced voltage is 12V

Explanation of Solution

Given:

  

  $\frac{d{i}_p}{dt}=0.3A/ms$

Formula used:

Primary induced voltage is calculated by

  $e_p=L_p\frac{d{i}_p}{dt}$

The secondary induced voltage is calculated by

  $e_s=M\frac{d{i}_p}{dt}$

Calculation:

With the known values, the primary induced voltage is calculated as

  $e_p=L_p\frac{d{i}_p}{dt}$

Substitute ${\text{L}}_p=50mH\&\frac{d{i}_p}{dt}=0.3A/ms$

  $e_p=(50mH)(0.3A/s)=15V$

Thus, the primary induced voltage is 15V

The secondary induced voltage is

  $e_s=M\frac{d{i}_p}{dt}$

Substitute $M=40mH\&\frac{d{i}_p}{dt}=0.3A/ms$

  $e_s=(40mH)(0.3A/s)=12V$

The secondary induced voltage is 12V

Conclusion:

Thus, the value of primary induced voltage is 15V and the value of secondary induced voltage is 12V