Question

Ip= 19A at Vp = 440V and Vs= 9000V. If the transformer is 95% efficient, how many turns are in the secondary coil? (efficiency e is the ratio of the power output to the power input or e = Ps/Pp). I know the result is Ns= (Vs Np) / Vp = 3680 turns, but why can't I use this below equation:

Ps/Pp = (Is Vs) / (Ip Vp) = 0.95

Is/Ip = Np/Ns

we have Np Vs / Ns Vp = 0.95

Ns= (Vs Np) / (0.95 Vp)

So why can't I use this equation [ Ns= (Vs Np) / (0.95 Vp) ] to find secondary coil.
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Could you please draw a diagram of this case. Please explain why I = 240V / (R + r) ? Where does r come from and what formula is this? a 240-V dc motor has an armature whose resistance is 1.8 ohms. When running at its operating speed, it draws a current of 14.0 A. Starting current is 133 A. What series resistance would be required to limit the starting current to 26.0 A? Could you also explain what is a back emf with simple definition? why does it act against applied voltage and reduce the current?

Answer 1

Soln :

Given Ip = 19A Vp= 440V Vs = 9000V, Transformerefficiency = 95% = 0.95

The transformer ratio can be given as Vs/Vp = Ns/ Np

For an Ideal tranformer, Input power = Output Power

ie, VpIp = VsIs ( Pp = Ps )

Therefore, Vs/Vp = Ip/Is = Ns/Np

Ns/ Np = Pp /Ps

Transformer efficiency = (Output power/Input power) x 100 =95% = 0.95

0.95 = VsIs/VpIp

ie NpVs/NsVp =0.95

Therefore Ns = NpVs/0.95Vp

Here this equation is not posssible since Np is not given.