Back emf and battery emf in RL Circuit

AI Thread Summary
In an RL circuit, the back EMF generated by the inductor cannot exceed the battery EMF. The effective voltage of the source is determined by the relationship between the voltages across the inductor and resistor, expressed as V_S = √(V_L² + V_R²). Without a capacitor in the circuit, the back EMF remains constrained by the battery's EMF. Therefore, it is concluded that back EMF cannot surpass the battery EMF in this configuration. Understanding these principles is essential for analyzing RL circuits effectively.
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hello all!:smile:

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?
 
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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.)
 
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