If the applied voltage of a DC motor is 230 V, then back emf, for maximum power developed is
A
460 V
B
230 V
C
200 V
D
115 V

Question

If the applied voltage of a DC motor is 230 V, then back emf, for maximum power developed is

Examoogle · Accepted Answer

115 V

Explanation: Given: Applied voltage V = 230 V, Power condition: Maximum power developed

Formula Used: $$E_b = \frac{V}{2}$$

Steps: Understanding the condition for maximum power: In a DC motor, the power developed is given by the product of back emf $E_b$ and armature current $I_a$. Given $V = E_b + I_a R_a$, the power $P = E_b I_a = E_b \left( \frac{V - E_b}{R_a} \right)$.. Applying calculus for maximum power: To find the maximum power, differentiate power $P$ with respect to $E_b$ and set it to zero: $\frac{dP}{dE_b} = \frac{V - 2E_b}{R_a} = 0$.. Solving for Back EMF: Solving the equation for $E_b$, we get $E_b = V / 2$. Substituting the given voltage $V = 230$ V, we find the back emf.

Answer: D is correct because for maximum power developed in a DC motor, the back emf must be exactly half of the applied terminal voltage, which is $115$ V.

Exam Tip: This derivation is analogous to the Maximum Power Transfer Theorem in circuit analysis, where load resistance must equal source resistance for maximum power output.

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