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

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

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