A dc motor with its rotor and field coils connected in series has an internal resistance of$\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\Omega }$. When the motor is running at full load on aline, the emf in the rotor is105 V. (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

Electric power is the rate at which work is done or energy is transformed into an electrical circuit.

V = E + lr

The value of internal resistance is,$\mathrm{r}=3.2\mathrm{\Omega }$ .

The emf of the rotor is,$\epsilon =105V$ .

The value of volt is,$V_{ab}=120V$ .

The sketch from the question is

The loop rule is

$V_{ab}=E+lr$

Or,

$I=\frac{V_{ab}-\mathrm{E}}{r}$

Substitute all value in the above equation.

$\begin{array}{r}I=\frac{V_{ab}-\mathrm{E}}{r}\\ =\frac{120\mathrm{V}-105\mathrm{V}}{3.2\mathrm{\Omega }}\\ =4.7\mathrm{A}\end{array}$

Therefore, 4.7 A current drawn by the motor from the line.

(b)

The power delivered to the motor line is,

${\mathrm{P}}_{\mathrm{in}}={\mathrm{IV}}_{\mathrm{ab}}$

Substitute all value in the above equation.

$\begin{array}{r}{P}_{\text{in}}=I{V}_{ab}\\ =\left(4.7\mathrm{A}\right)\left(120\mathrm{V}\right)\\ =564\mathrm{W}\end{array}$

Therefore, the power delivered to the motor is 564 W

(c)

The mechanical power developed by the motor is,

$\begin{array}{r}{P}_{\text{out}}=P_{\text{in}}-P_{\text{loss}}\\ P_{\text{out}}=I{V}_{ab}-I^{2}r\end{array}$

Substitute all value in the above equation.

$\begin{array}{r}{P}_{\text{out}}=564\mathrm{W}-\left(4.7\mathrm{A}{)}^2\right(3.2\mathrm{\Omega })\\ =493\mathrm{W}\end{array}$

Therefore, 493 W mechanical power developed by the motor.