Question

A 5 kW, 220 V DC shunt motor has \( 0.5 \, \Omega \) armature resistance including brushes. The motor draws a no-load current of \( 3 \, A \). The field current is constant at \( 1 \, A \). Assuming that the core and rotational losses are constant and independent of the load, the current (in amperes) drawn by the motor while delivering the rated load, for the best possible efficiency, is (rounded off to 2 decimal places).

Hint:

The total current in a DC shunt motor is the sum of the armature current and the field current.

Answer 1

Step 1: Calculate the full-load current. The rated power of the motor is: Case 1: At no load The generated EMF (\( E_b \)) is given by: \[ E_b = V - I_a R_a \] Substitute the values: \[ E_b = 220 - (3 \cdot 1) \cdot 0.5 = 219 \, \text{Volts}. \] The no-load power developed (equal to mechanical loss): \[ \text{No-load power} = E_b I_a \] Substitute the values: \[ \text{No-load power} = 219 \cdot 2 = 438 \, \text{W}. \] Case 2: At rated load Given: \[ P_{\text{out}} = 5 \, \text{kW} = 5000 \, \text{W}. \] Mechanical power developed is equal to the net mechanical power output plus mechanical losses: \[ E_b I_a = 5000 + 438 = 5438 \, \text{W}. \] The generated EMF equation: \[ (V - I_a R_a) I_a = 5438 \] Substitute the values: \[ (220 - I_a \cdot 0.5) I_a = 5438 \] Rearrange the equation: \[ 0.5 I_a^2 - 220 I_a + 5438 = 0 \] Solve the quadratic equation for \( I_a \): \[ I_a = 26.28 \, \text{A}. \] The total current drawn by the motor is: \[ I_L = I_a + I_{\text{sh}} \] Substitute \( I_a = 26.28 \, \text{A} \) and \( I_{\text{sh}} = 1 \, \text{A} \): \[ I_L = 26.28 + 1 = 27.28 \, \text{A}. \] Final Answer: The current drawn by the motor is: \[ I_L = 27.28 \, \text{A}. \]