Question
Find the armature current in a dc shunt motor if $E_a=300 \mathrm{~V}$ and the motor develops $4 \mathrm{~kW}$ of power.
Step 1
We have the following information: - Armature voltage, \( E_a = 300 \, \text{V} \) - Power developed by the motor, \( P = 4 \, \text{kW} = 4000 \, \text{W} \) Show more…
Show all steps
Your feedback will help us improve your experience
Narayan Hari and 83 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A certain de motor has $R_{A}=1.3 \Omega, I_{A}=$ $10 \mathrm{A},$ and produces a back emf $E_{A}=$ $240 \mathrm{V},$ while operating at a speed of 1200 rpm. Determine the voltage applied to the armature, the developed torque, and the developed power
A motor has a back emf of $110 \mathrm{~V}$ and an armature current of 90 A when running at 1500 rpm. Determine the power and the torque developed within the armature. Power $=($ Armature current $)($ Back emf $)=(90 \mathrm{~A})(110 \mathrm{~V})=9.9 \mathrm{~kW}$ From Chapter 10, power $=\tau \omega$ where $\omega=2 \pi f=2 \pi(1500 \times 1 / 60)$ $\mathrm{rad} / \mathrm{s}$ $$ \text { Torque }=\frac{\text { Power }}{\text { Angular speed }}=\frac{9900 \mathrm{~W}}{(2 \pi \times 25) \mathrm{rad} / \mathrm{s}}=63 \mathrm{~N} \cdot \mathrm{m} $$
Some generators, called shunt generators, use electromagnets in place of permanent magnets, with the field coils for the electromagnets activated by the induced voltage. The magnet coil is in parallel with the armature coil (it shunts the armature). As shown in Fig. $33-3$, a certain shunt generator has an armature resistance of $0.060 \Omega$ and a shunt resistance of $100 \Omega$. What power is developed in the armature when it delivers $40 \mathrm{~kW}$ at $250 \mathrm{~V}$ to an external circuit? From $\mathrm{P}=V I$ $$ \begin{array}{l} \text { Current to the external circuit }=I_{x}=\frac{\mathrm{P}}{V}=\frac{40000 \mathrm{~W}}{250 \mathrm{~V}}=160 \mathrm{~A} \\ \text { Field current }=I_{f}=\frac{V_{f}}{r_{f}}=\frac{250 \mathrm{~V}}{100 \Omega}=2.5 \mathrm{~A} \\ \text { Armature current }=I_{a}=I_{x}+I_{f}=162.5 \mathrm{~A} \end{array} $$ Total induced emf $=|\mathscr{E}|=\left(250 \mathrm{~V}+I_{a} r_{a}\right.$ drop in armature $$ \begin{aligned} &=250 \mathrm{~V}+(162.5 \mathrm{~A})(0.06 \Omega)=260 \mathrm{~V} \\ \text { Armature power } &=I_{a}|\mathscr{E}|=(162.5 \mathrm{~A})(260 \mathrm{~V})=42 \mathrm{~kW} \end{aligned} $$ $$ \begin{array}{l} \text { Alternative Method }\\ \begin{array}{l} \text { Power loss in the armature }=I_{a}^{2} r_{a}=(162.5 \mathrm{~A})^{2}(0.06 \Omega)=1.6 \mathrm{~kW} \\ \qquad \begin{aligned} \text { Power loss in the field } &=I_{f}^{2} r_{f}=(2.5 \mathrm{~A})^{2}(100 \Omega)=0.6 \mathrm{~kW} \\ \text { Power developed } &=(\text { Power delivered })+(\text { Power loss in armature })+(\text { Power loss in field) }\\ &=40 \mathrm{~kW}+1.6 \mathrm{~kW}+0.6 \mathrm{~kW}=42 \mathrm{~kW} \end{aligned} \end{array} \end{array} $$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD