Question

You are an electrician on the job. It is decided that the speed of a large DC motor is to be reduced by connecting a resistor in series with its armature. The DC voltage applied to the motor is 250 V, and the motor has a full-load armature current of 50 A. Your job is to reduce the armature current to 40 A at full load bu connecting the resistor in series with the armature. What value of resistance should be used, and what is the power rating of the resistor?

Answer 1

The equivalent resistance of a series circuit is found by finding the sum of the individual resistances

The series resistor required is 1.25 Ω

The power rating of the resistor, is 2,000 W

The reason why the above values are correct are as follows;

The given parameter are;

Applied DC Voltage, V = 250 V

The full load armature current, I = 50 A

Required reduced armature current, $I_r$ = 40 A

Solution:

First part

The resistance, R, of the circuit, is given as follows;

$R = \dfrac{V}{I}$

Therefore;

$R = \dfrac{250 \, V}{50 \, A} = 5 \, \Omega$

The voltage, V = I·R, therefore, with a series resistor, $R_s$, we can have;

V = I·R = $I_r$·($R_s$ + R) (given that the same current flows through both resistor)

Where the required current is $I_r$ = 40 A, we get;

$R_s = \dfrac{I \cdot R}{I_r} - R$

Which gives;

$R_s = \dfrac{50 \times 5}{40} - 5 = 1.25$

The required series resistor, $R_s$ = 1.25 Ω

Second part;

Power, P = I²·R

The power rating of the resistor, which is added to the circuit to regulate the current to 40A is therefore, given as follows;

$P_r$ = $I_r$²·$R_s$

∴ $P_r$ = 40² × 1.25 = 2,000

The power rating of the resistor is 2,000 W

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