An inductor of inductance L=400\: mH and resistors of resistances R_{1}=2\, \Omega \; and\; R_{2}=2\, \Omega are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t=0 . The potential drop across L as a function of time is

  • Option 1)

    6e^{-5t}V\; \;

  • Option 2)

    \; \frac{12}{t}e^{-3t}V\;

  • Option 3)

    \; \; 6(1-e^{-t/0.2})V\;

  • Option 4)

    \; 12e^{-5t}V

 

Answers (1)

As we learnt in 

Self Inductance -

An emf is induced in the coil or the circuit which oppose the change that causes it. Which is also known back  emf.

- wherein

 

 time constant of the given circuit isT=\frac{\mathrm{L} }{\mathrm{R} _{2}}=\frac{400mh}{2}=0.2 sec

\therefore potential drop is

\xi =\varepsilon _{0}\cdot e^{-}\frac{\mathrm{t} }{\mathrm{T} } \\\xi =12\cdot e^{-5t}


Option 1)

6e^{-5t}V\; \;

This is incorrect

Option 2)

\; \frac{12}{t}e^{-3t}V\;

This option is incorrect

Option 3)

\; \; 6(1-e^{-t/0.2})V\;

This option is incorrect

Option 4)

\; 12e^{-5t}V

This option is correct

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