A steady voltage is applied to a dc motor. The armature winding resistance is equal to R. The maximum useful power of the motor is

\frac{V^{2}}{R}

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\frac{V^{2}}{2 R}

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\frac{V^{2}}{4 R}

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\frac{3 V^{2}}{4 R}

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Solution

The correct option is C $\frac{V^{2}}{4 R}$
In dc motor, current through the armature winding $I = \frac{V - ε}{R}$ where $ε$ is the back emf

For maximum efficiency, $ε = \frac{V}{2}$ so, $I = \frac{V}{2 R}$

Power $= I^{2} R = \frac{V^{2}}{4 R}$

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A steady voltage is applied to a dc motor. The armature winding resistance is equal to R. The maximum useful power of the motor is

The correct option is C V24RIn dc motor, current through the armature winding I=V−εR where ε is the back emfFor maximum efficiency, ε=V2 so, I=V2RPower =I2R=V24R

The correct option is C $\frac{V^{2}}{4 R}$
In dc motor, current through the armature winding $I = \frac{V - ε}{R}$ where $ε$ is the back emf

For maximum efficiency, $ε = \frac{V}{2}$ so, $I = \frac{V}{2 R}$

Power $= I^{2} R = \frac{V^{2}}{4 R}$