Answer:
I. Vs = 8.0 Volts.
II. Is = 4.5 Amperes.
Explanation:
Given the following data;
Np = 1.5Ns = [tex] \frac {N_{P}}{N_{S}} = 1.5 [/tex] ..... equation 1
Ip = 3.0 A
Vp = 12 V
To find the voltage and current on the secondary coil;
I. For the voltage in the secondary coil (Vs), we would use the following formula;
[tex] \frac {V_{P}}{V_{S}} = \frac {N_{P}}{N_{S}} [/tex] ...... equation 2.
Substituting eqn 1 into eqn 2, we have;
[tex] \frac {V_{P}}{V_{S}} = 1.5 [/tex]
[tex] \frac {12}{V_{S}} = 1.5 [/tex]
Cross-multiplying, we have;
[tex] V_{S} * 1.5 = 12 [/tex]
[tex] V_{S} = \frac {12}{1.5} [/tex]
Vs = 8.0 V
II. For the current in the secondary coil (Is), we would use the following formula;
[tex] \frac {I_{S}}{I_{P}} = \frac {N_{P}}{N_{S}} [/tex] .... equation 3
Substituting eqn 1 into eqn 3, we have;
[tex] \frac {I_{S}}{I_{P}} = 1.5 [/tex]
[tex] \frac {I_{S}}{3.0} = 1.5 [/tex]
Cross-multiplying, we have;
[tex] I_{S} = 1.5 * 3.0 [/tex]
Is = 4.5 A
