Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Table of contents
Textbook solutions
All Solutions
Page 684: Practice Problems
Exercise 16
Step 1
1 of 2
In order to solve this problem, we are going to use the transformer equation which is given as
$frac{I_s}{I_p}=frac{V_p}{V_s}=frac{N_p}{N_s}$
Now, we can express the voltage of the secondary circuit
$frac{I_s}{I_p}=frac{V_p}{V_s}=frac{N_p}{N_s}$
Now, we can express the voltage of the secondary circuit
$$
V_s=frac{N_s}{N_p}V_p=frac{125}{7500}times 7200
$$
Which gives that
$$
boxed{V_s=120textrm{ V}}
$$
Now, we can express the current in the primary circuit
$$
I_p=frac{N_s}{N_p}I_s=frac{125}{7500}times 36
$$
$$
boxed{I_p=0.6textrm{ A}}
$$
Result
2 of 2
$$
V_s=120textrm{ V}
$$
V_s=120textrm{ V}
$$
$$
I_p=0.6textrm{A}
$$
Exercise 17
Step 1
1 of 2
In order to solve this problem, we are going to use the transformer equation which is given as
$frac{I_s}{I_p}=frac{V_p}{V_s}=frac{N_p}{N_s}$
Now, we can express the voltage of the secondary circuit
$frac{I_s}{I_p}=frac{V_p}{V_s}=frac{N_p}{N_s}$
Now, we can express the voltage of the secondary circuit
$$
V_s=frac{N_s}{N_p}V_p=frac{90000}{300}times 60
$$
Which gives that
$$
boxed{V_s=18times 10^3textrm{ V}}
$$
Now, we can express the current in the primary circuit
$$
I_p=frac{N_s}{N_p}I_s=frac{90000}{300}times 0.5
$$
$$
boxed{I_p=150textrm{ A}}
$$
Result
2 of 2
$$
V_s=18times 10^3textrm{ V}
$$
V_s=18times 10^3textrm{ V}
$$
$$
I_p=150textrm{ A}
$$
unlock