# DC Motor Voltage Equation, Power Equation and Condition for Maximum Mechanical Power

Electronics & ElectricalElectronDigital Electronics

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## DC Motor Voltage Equation

Consider a DC shunt motor as shown in the figure.  Let,

$$\mathrm{V\:=\:Applied\:DC\:voltage}$$

$$\mathrm{E_{b}\:=\:Back\:EMF}$$

$$\mathrm{R-{a} \:=\:Resistance\:of\:the\:armature\:circuit}$$

$$\mathrm{I_{a}\:=\:Armature\:current}$$

By applying the KVL in the armature circuit, we get,

$$\mathrm{V\:=\:E_{b}\:+\:I_{a}R_{a} … (1)}$$

The expression in eqn. (1) is known as the voltage equation of the DC motor.

## Power Equation of DC Motor

If the eqn. (1) is multiplied by Ia on both sides, we obtain,

$$\mathrm{VI_{a}\:=\:E_{b}I_{a}\:+\:I_a^2R_{a}\:.... (2)}$$

The eqn. (2) is known as the power equation of DC motor. The mechanical losses are neglected.

Where,

$$\mathrm{VI_{a}\:=\:Input\:electrical\:power\:to\:the\:armature}$$

$$\mathrm{E_{b}I_{a}\:=\:Electromechanical\:power\:developed\:by\:armature}$$

$$\mathrm{I_a^2R_{a}\:=\:Electric\:power\:wasted\:in\:the\:armature\:Cu\:losses}$$

## Condition for Maximum Mechanical Power

The mechanical power developed by the motor (neglect mechanical losses) is given by,

$$\mathrm{P_{m}\:=\:E_{b}I_{a}}$$

$$\mathrm{⇒\:P_{m}\:= VI_{a}I_a^2R_{a}}$$

Since, the V and Ra are constant for a given machine, hence the mechanical power developed by the motor depends upon the armature current.

The condition for maximum power is given by,

$$\mathrm{\frac{dP_{m}}{dI_{a}}\:=\:0}$$

$$\mathrm{\frac{d}{dI_{a}}(VI_{a}-I_a^2R_{a})\:=\:0}$$

$$\mathrm{⇒\:V\:-\:2I_{a}R_{a}\:=\:0}$$

$$\mathrm{⇒\:I_{a}R_{a\:=\:\frac{V}{2}}\:......(4)}$$

Therefore,

$$\mathrm{V\:=\:E_{b}\:+\:I_{a}R_{a}\:=\:E_{b}\:+\:\frac{V}{2}}$$

$$\mathrm{⇒E_{b}\:=\:\frac{V}{2}\:.....(5)}$$

Hence, the mechanical developed by a DC motor being maximum, when the back EMF is equal to the half of the applied voltage.