Respuesta :
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, [tex]I_r[/tex] = 40 A
Solution:
First part
The resistance, R, of the circuit, is given as follows;
[tex]R = \dfrac{V}{I}[/tex]
Therefore;
[tex]R = \dfrac{250 \, V}{50 \, A} = 5 \, \Omega[/tex]
The voltage, V = I·R, therefore, with a series resistor, [tex]R_s[/tex], we can have;
V = I·R = [tex]I_r[/tex]·([tex]R_s[/tex] + R) (given that the same current flows through both resistor)
Where the required current is [tex]I_r[/tex] = 40 A, we get;
[tex]R_s = \dfrac{I \cdot R}{I_r} - R[/tex]
Which gives;
[tex]R_s = \dfrac{50 \times 5}{40} - 5 = 1.25[/tex]
The required series resistor, [tex]R_s[/tex] = 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;
[tex]P_r[/tex] = [tex]I_r[/tex]²·[tex]R_s[/tex]
∴ [tex]P_r[/tex] = 40² × 1.25 = 2,000
The power rating of the resistor is 2,000 W
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