# Exam Help

Discussion in 'Physics' started by Hello, Dec 26, 2008.

1. ### Hello Thread Starter Active Member

Dec 18, 2008
82
0
Can you please have look at the attached file.
I am having difficulty with Q6 part (v), not sure how to go about it!

Torque constant, Kt = 0.8
Back e.m.f constant, Ke = 1.25

Any help is greatly appreciated.

• ###### DEN109.pdf
File size:
255 KB
Views:
54
Last edited: Dec 26, 2008
2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You know Ke, thus

The voltage developed across the motor terminals, without the 5 ohm resistor connected, is

E=Ke*ω

You know Ke and ω thus you can calculate E (which is actually the back EMF when it is working as a motor).

The armature resistance is 5 ohms and the load resistance (connected on the motor terminals) is also 5 ohms. Thus the total resistance is 10 ohms. Thus the current through the armature and the load resistance is :

I=E/10

Thus the power dissipated in the load is:

P=(I^2)*5

Assuming no losses, the torque is:

T=P/ω

3. ### Hello Thread Starter Active Member

Dec 18, 2008
82
0
Thanks for your help, very well explained.

4. ### Hello Thread Starter Active Member

Dec 18, 2008
82
0
Im not very sure what 'load resistance' is ?
If possible could you go to a bit more detail.

Thanks

5. ### mik3 Senior Member

Feb 4, 2008
4,846
63
The load resistance is just a resistor, 5 ohms in this case, which is connected across the terminals of the motor.

6. ### Hello Thread Starter Active Member

Dec 18, 2008
82
0
Thanks, great help!