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 $P = E_b I_a = E_b \left( \frac{V - E_b}{R_a} \right)$.

$P = \frac{V E_b - E_b^2}{R_a}$

Applying calculus for maximum power

To find the maximum power, differentiate power $P$ with respect to $E_b$ and set it to zero: $\frac{dP}{dE_b} = \frac{V - 2E_b}{R_a} = 0$.

$V - 2E_b = 0$

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.

$E_b = \frac{230}{2} = 115 \text{ V}$