A transformer has 480 primary turns and 7.8 secondary turns. (a) If Vp is 120 V (rms), what is Vs with an open circuit? If the secondary now has a resistive load of 17 Ω, what is the current in the (b) primary and (c) secondary?

a) 1.95 V

b) 1.87 mA

c) 0.115 A

Explanation:

Given that

Number of primary turns, N(p) = 480

Number of secondary turns, N(s) = 7.8

Velocity of primary turns, V(p) = 120 V

Velocity of secondary turns, V(s) = ?

Current in the primary, I(p) = ?

Current in the secondary, I(s) ?

To solve this question, we would be using the formula

V(s)/V(p) = N(s)/N(s), now substituting the values, we have

V(s) / 120 = 7.8 / 480

V(s) = (7.8 * 120) / 480

V(s) = 936 / 480

V(s) = 1.95 V

To find the current in the primary, remember ohms law?

I = V/R

I(s) = V(s) / R(s)

I(s) = 1.95 / 17

I(s) = 0.115 A

Now, remember the relationship between current and voltage

I(p)/I(s) = V(s)/V(p)

I(p) / 0.115 = 1.95 / 120

I(p) = (1.95 * 0.115) / 120

I(p) = 0.22425 / 120

I(p) = 0.00187 A

I(p) = 1.87 mA

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5