# Power Equation of a DC motor

Voltage equation of a D.C. Motor

In case of a generator, generated emf has to supply armature resistance drop and remaining part is available across the load as a terminal voltage. But in case of d.c. motor, supply voltage V has to overcome back emf Eb which is opposing V and also various drops as armature resistance drop IaRa, brush drop etc.

Hence the voltage equation of a d.c. motor can be written as,

$\displaystyle V = E_b + I_a R_a + V_{br}$

$\displaystyle V_b \rightarrow$ Brush drop

Neglecting the brush drop, the generalized voltage equation is,

In case of Motor $\displaystyle \Rightarrow V = E_b + I_a R_a$

In case of Generator $\displaystyle \Rightarrow V = E_b - I_a R_a$

In case of motor, the back emf is always less than the supply voltage. But $\displaystyle R_a$ is very small hence under normal running conditions, the difference between back emf and the supply voltage is very small. The net voltage across the armature is the difference between the supply voltage and back emf which decides the armature current. Hence from the voltage equation we can write,

$\displaystyle I_a = \biggl ( \frac{V-E_b}{R_a} \biggr )$

Power Equation of a D.C. Motor

The voltage equation of a d.c. motor is given by,

$\displaystyle V = E_b + I_a R_a$

Multiplying both sides of the above equation by $\displaystyle I_a$ we get,

$\displaystyle VI_a = I_a E_b + I^2_a R_a$

This equation is called power equation of a d.c. motor.

$\displaystyle VI_a$ = Net electrical power input to the armature measured in watts.

$\displaystyle I^2_a R_a$ = Power loss due to the resistance of the armature called armature copper loss.

So difference between $\displaystyle VI_a$ and $\displaystyle I^2_a R_a$ i.e. input – losses gives the output power.

So $\displaystyle E_b I_a$ is called electrical equivalent of gross mechanical power developed by the armature. This is denoted as $\displaystyle P_m$

Gross mechanical power developed in the armature = Power input to the armature – Armature copper loss

Condition for Maximum power

For a motor from power equation it is known that

$\displaystyle P_m = VI_a - I^2_a R_a$

For maximum gross mechanical power

$\displaystyle \Rightarrow \frac {dP_m}{dI_a} = 0$

$\displaystyle V - 2 I_a R_a = 0$

$\displaystyle I_a = \frac {V}{2R_a}$

$\displaystyle I_a R_a = \frac {V}{2}$

$\displaystyle V = E_b + I_a R_a = E_b + \frac {V}{2}$

$\displaystyle E_b = \frac {V}{2}$