91Ó°ÊÓ

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,$\mathrm{r}=3.2\mathrm{Î\copyright }$ .

The emf of the rotor is,$\mathrm{Î\mu }=105V$ .

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.

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

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.

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

Therefore, 493 W mechanical power developed by the motor.