Question

A 25 kW, 250 V dc machine is separately excited. The filed current is held constant at a speed of 3000 rpm. The open circuit voltage is 250 V. The terminal voltage is 255V. Armature resistance is $0.05\Omega$. The electromagnetic torque developed will be

• A. 25 N-m
• B. 141. 46 N-m
• C. 38.74 N-m
• D. 79.58 N-m

Answer 1

The correct option is D 79.58 N-m

Open circuit $I_a=0$, then
$V_t=E_a=250V$ at $3000rpm$

Now, $\begin{array}{cc}{V}_t& =255V\end{array}$
As $V_t>E_a$ the machine is acting as a motor
$\begin{array}{cc}{I}_a& =\frac{V_t-E_a}{R_a}=\frac{255-250}{0.05}=100A\end{array}$
The current flowing into the positive terminal in opposition to $E_a$
$\therefore$ Electromagnetic power
$=E_a{I}_a=250\times 100$
$=25kW=$ mechanical power output

Speed $=3000$ rpm
or, $\frac{3000\times 2\pi }{60}=314.16rad/s$
Electromagnetic torque,
$T_{em}=\frac{E_a{I}_a}{{\omega }_m}=\frac{25\times {10}^3}{314.16}=79.58N-m$