The armature of a DC motor has resistance of 0.1 ohm and is connected to a 250 V supply. The generated emf when the motor is taking 60 A will be ____.
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Remember: In a Motor, the generated emf ($E_b$) is always less than the supply voltage because it opposes the flow of current. In a Generator, the generated emf ($E_g$) is always greater than the terminal voltage.
Step 1: Find the total electrical power input to the motor: \( P_{in} = V \times I_a = 250 \times 60 = 15000\text{ W} \).
Step 2: Find the power wasted as heat in the armature resistance: \( P_{loss} = I_a^2R_a = 60^2 \times 0.1 = 3600 \times 0.1 = 360\text{ W} \).
Step 3: The remaining power is what's actually converted at the back emf; dividing this developed power by the current gives \( E_b \): \( P_{dev} = P_{in} - P_{loss} = 15000 - 360 = 14640\text{ W} \), and \( E_b = \frac{P_{dev}}{I_a} = \frac{14640}{60} \).
\[ \boxed{E_b = 244 \text{ V}} \]