Question

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?

Short answer

(a) 4.7 A Current drawn by the motor from the line.

(b) The power delivered to the motor is 564 W

(c) 493 W Mechanical power developed by the motor.

Step-by-step solution

Electric power

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

V = E + lr

Step 2: Identification of the given data

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,$V_{ab}=120V$ .

Determine the current drawn by motor line

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.

 Step 3: Determine the power delivered to the motor

(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

Determine the mechanical power developed by the motor

(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.