Part A Exercise 27.48 What is the current drawn by the motor from the line? Expr

Question

Part A Exercise 27.48 What is the current drawn by the motor from the line? Express your answer using two significant figures Adc motor with its rotor and field coils connected in series has an internal resistance of 3.2 . When running at full load on a 120-V line, the emf in the rotor is 105 V Submit My Answers Give Up Part B What is the power delivered to the motor? Express your answer using two significant figures. Submit My Answers Give Up Part C What is the mechanical power developed by the motor? Express your answer using two significant figures. Submit My Answers Give Up

Explanation / Answer

Let the internal resistance be r = 3.2 ohm.

Input voltage V' = 120 V

Emf in the rotor is given by E = 105 V

a) Let the current drawn from the line be I

The input voltage is equal to sum of emf of rotor and internal voltage ( or voltage across the series resistor ).

V' = E + Ir  

120 = 105 + ( I * 3.2 )

I = ( 120 - 105 ) / 3.2

Current I = 4.6875 A

b) Power delivered to motor = Input voltage × Current

P = 120 × 4.6875 = 562.5 W

c) To know the power delivered by the motor , we need to calculate the power dissipated in the motor.

Power dissipated Pd = I2 r

= 4.6875 × 4.6875 × 3.2

= 70.3125 W

We can also calculate this by taking voltage across the resistor as resistor is the one which dissipates the energy.

Voltage across the resistor is V'' = V' - E = 120 - 105 = 15 V

( or ) I r = V' - E = 120 - 105 = 15 V

Again Pd = V''2 / r

= (15 × 15 ) / 3.2

= 225 / 3.2

= 70.3125 W

Now the power delivered by the motor is

P'' = P - Pd

= 562.5 - 70.3125

= 492.1875 W

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