DC Motor Voltage, Power and Max Power Condition

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DC Motor Voltage Equation

Consider a DC shunt motor as shown in the figure.

Let,

V = Applied DC voltage

E_b = Back bF

R_a = Resistance of the Armature Circuit

I_a = Armature Current

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

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

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 I_a on both sides, we obtain,

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

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

Where,

VI_a = Input electrical power to the armature

E_bI_a = Electromechanical power developed by armature

I_a²R_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,

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

$$\mathrm{\Rightarrow \:P_{m}\:=\: VI_{a}I_{a}^{2}R_{a}}$$

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,

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

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

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

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

Therefore,

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

$$\mathrm{\Rightarrow \:E_{b}\:=\:\frac{V}{2}\:\: \dotso \:(5)}$$

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