DC Motor Voltage Equation, Power Equation and Condition for Maximum Mechanical Power
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.
Published on 13-Aug-2021 10:54:16
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