Question

Suppose a motor connected to a 135 V source draws 21.5 A when it first starts.

(a) What is its resistance (in Ω)? Answer in Ω.

(b) What current (in A) does it draw at its normal operating speed when it develops a 100 V back emf? Answer in A.

(c) What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 525 turns, its secondary 8 turns, and the input voltage is 112 V?

(d) What input current (in mA) is required to produce a 4.00 A output?

(e) What is the power input (in W)?

Answer 1

The supply voltage is V = 135 V and , Current at first Io = 21.5 A

a ) the resistance R = V oltage / current = 135 / 21.5 = 6.28 ohm

b) the back emf Vb = 100 V , hence the net Voltage is Vnet = 135- 100 V = 35 V

Vnet = 35 V , and resistance R = 6.28 Ohm ,

hence current is I = V/R = 35/6.28 = 5.57 A

c) The ratio of no of turns and the voltage is given by

where

Ns is the no . of turns of secondary coil = 8

Np is the no . of turns of primary coil = 525

Vp voltage at primary (input) = 112 V

Vs voltage at secondary (output )  

putting these values we have

so Vs = 1.7 V

d) For an ideal transformer the input power is equal to the output power so

Vin = 112 V , Vout = 1.7 V , I out = 4 A so

so , 112 X = 1.7 X 4  

or = 0.0607 A or 60.7 mA

e) The input power is