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  • Shown in the figure is a R-L circuit. Just after the Key (K) is closed <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/07/phy-q4-file21.PNG" style="height:107px; width:

Shown in the figure is a R-L circuit. Just after the Key (K) is closed

Option: 1

the current in the circuit is zero


Option: 2

potential drop across the resistor is zero


Option: 3

emf develop across the inductor equals the emf of the battery


Option: 4

ALL of above options are correct


Answers (1)

best_answer

If the equilibrium current in the circuit is i0 , current as any time in transient state is given as

\mathrm{i=i_0\left(1-\frac{f^{\frac{t}{\tau}}}{\mathrm{e}}\right) where \ \tau=\frac{\mathrm{L}}{\mathrm{R}} }        \mathrm{\text { Putting } \mathrm{t}=0, \mathrm{i}=\mathrm{i}_0=0}

= The voltage drop across the inductor is \mathrm{E^{\prime}} that oppose the applied emf E

So  \mathrm{E^{\prime}} = E numerically.

Since the circuit current is zero, the heat loss that is proportional to R will be zero Hence all the statements are correct

 

 

Posted by

Rishi

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