Back emf and battery emf in RL Circuit

[k5591002]

hello all!

I have a question about electromagnetism.
The question is "Can the back emf ever be greater than the battery emf in RL circuit?"

can someone explain this one for me please?

 

[Chen]

I'm assuming the resistor R and the inductor L are connected in series. As you know the effective V of the source in RLC circuits like this is given by:

$$V_S = \sqrt{(V_L - V_C)^2 + V_R^2}$$
(all voltages are effective)

In this case however, we don't have a capacitor so it's actually just:

$$V_S = \sqrt{V_L^2 + V_R^2}$$

With some basic operations you can find that:

$$V_L = \sqrt{V_S^2 - V_R^2}$$

And therefore the answer is No, the back EMF on the inductor cannot be greater than the EMF of the battery. (This is because there is no capacitor in the circuit, if there was one then it would be possible for the back EMF of the inductor to outgrow the EMF of the battery.)

 

[mmwave]

Originally posted by k5591002
hello all!

I have a question about electromagnetism.
The question is "Can the back emf ever be greater than the battery emf in RL circuit?"

can someone explain this one for me please?

I'm not sure of your definition of back emf but one thing novice designers of inductive circuits learn the hard way is that when you interrupt the current in an inductor (say by opening a relay or turning off the base drive to a transistor) the inductor will generate whatever voltage is necesary to maintain that current until the energy stored in the mag. field is gone. That's why inductors in switching circuits have a diode across them, to give a path for the current when you turn the transistor off (or open a switch). Putting in a larger transistor just makes for larger fireworks.