Back emf and battery emf in RL Circuit

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SUMMARY

The discussion clarifies that in an RL circuit, the back electromotive force (EMF) cannot exceed the battery EMF. The effective voltage of the source in such circuits is defined as V_S = √(V_L² + V_R²), where V_L represents the voltage across the inductor and V_R the voltage across the resistor. Since there is no capacitor present in this configuration, the back EMF generated by the inductor remains less than or equal to the battery EMF, confirming that the back EMF cannot surpass the battery EMF.

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  • Understanding of RL circuit configurations
  • Knowledge of effective voltage calculations in AC circuits
  • Familiarity with the concepts of back EMF and inductance
  • Basic principles of electromagnetism
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  • Study the role of capacitors in RLC circuits and their effect on back EMF
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Electrical engineers, physics students, and anyone interested in understanding the dynamics of RL circuits and the relationship between back EMF and battery EMF.

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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:

[tex]V_S = \sqrt{(V_L - V_C)^2 + V_R^2}[/tex]
(all voltages are effective)

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

[tex]V_S = \sqrt{V_L^2 + V_R^2}[/tex]

With some basic operations you can find that:

[tex]V_L = \sqrt{V_S^2 - V_R^2}[/tex]

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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