Question

In case of dc motor, maximum mechanical power is developed when back emf equals:
1. The applied voltage
2. Half the applied voltage
3. One third of the applied voltage
4. Double the applied voltage

Answer 1

Correct Answer - Option 2 : Half the applied voltage

For a dc motor from the power equation, it is known that,

Pm = Gross mechanical power developed = EbIa = VIa - Ia2Ra

Where,

V = Applied voltage

Eb = Back emf

Ia = Armature current

Ra = Armature resistance

For maximum Pm,

\(\begin{array}{l} \frac{{d{P_m}}}{{d{I_a}}} = 0\\ \Rightarrow 0 = V - 2{I_a}{R_a}\\ \Rightarrow {I_a}{R_a} = \frac{V}{2} \end{array}\)

Substituting in voltage equation (V = Eb + IaRa), we get

\(\begin{array}{l} V = {E_b} + {I_a}{R_a} = {E_b} + \left( {\frac{V}{2}} \right)\\ \Rightarrow {E_b} = \frac{V}{2} \end{array}\)

So that when a DC motor generates maximum power, ratio of applied voltage to back emf is 2 : 1