Answer :
The magnitude of the back e.m.f. at full load for the given DC shunt motor is 663.6 V.
A 230 V D.C. shunt motor takes 32 A at full load. Back e.m.f. formula for a DC motorE_b = V - Ia(R_a + R_s)Where,E_b is the back e.m.f.V is the applied voltage R_a is the resistance of the armature circuitR_s is the resistance of the shunt field windingI_a is the current through the armature circuitAt full load, the current through the armature circuit, I_a = 32 AThe resistance of the armature circuit, R_a = 0.2 ohmsThe resistance of the shunt field winding, R_s = 115 ohmsThe applied voltage, V = 230 VBack e.m.f. at full load can be found asE_b = V - I_a(R_a + R_s) = 230 - 32(0.2 + 115)≈ -663.6VThe negative sign indicates that the back e.m.f. opposes the applied voltage.
Hence, the magnitude of the back e.m.f. at full load for the given DC shunt motor is 663.6 V.
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Full-load 230 V D.C. shunt motors draw 32 A. The back e.m.f.(electromotive force) at full load for the motor is approximately -3,456.4 V.
To find the back electromotive force (e.m.f.) at full load for a 230 V D.C. shunt motor, we can use the equation:
E = V - I × (Rarm + Rshunt)
Where:
E is the back e.m.f.
V is the supply voltage.
I is the current at full load.
Rarm is the resistance of the motor armature.
Rshunt is the resistance of the shunt field winding.
Substituting the values into the equation:
E = 230 - 32 × (0.2 + 115)
E = 230 - 32 × 115.2
E = 230 - 3,686.4
E ≈ -3,456.4 V
The back e.m.f. at full load for the motor is approximately -3,456.4 V. Note that the negative sign indicates that the back e.m.f. opposes the applied voltage.
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