An electric motor is driven by d.c. supply of emf \( V_0 \) and internal resistance \( r \). The current when output power is maximum will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( V_0/2r \)
The power supplied by the source is \( V_0 I \); part of it is lost as heat \( I^{2}r \) in the internal resistance, and the rest is the output power of the motor.
\( \displaystyle P_{\mathrm{out}} = V_0 I - I^{2} r \)
For a maximum, the derivative with respect to \( I \) is zero.
\( \displaystyle \frac{dP_{\mathrm{out}}}{dI} = V_0 - 2Ir = 0 \)
\( \displaystyle I = \frac{V_0}{2r} \)
At this current the output power is \( V_0^{2}/4r \), and the back emf of the motor is \( V_0/2 \).