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An emf of 20 V is applied at time t =0 to a circuit containing in series 10mH inductor and 5\: \Omega resistor. The ratio of the currents at time t=\infty and at t=40\: s is close to: (Take\; e^{2}=7.389)
 
Option: 1 1.15
Option: 2 1.06
Option: 3 0.84
Option: 4 1.46
 

Answers (1)

best_answer

 

 

 

 

  

As we know that the -

    i = i_o(1-e^{\frac{-t}{L/R}})

        

 

So, from the question, we can write that -

     i = \frac{20}{5}(1-e^{\frac{-t}{0.01/5}})

  = i = 4(1-e^{ { -500t } })

By putting t = \infty

We get, i_o = 4

Again at t = 40 sec

 

    i_{40}= 4(1-\frac{1}{(e^2)^{10000}})

  = 4(1-\frac{1}{(7.389)^{10000}})

So,

\frac{i_{\infty}}{i_{40}} \approx 1.15

So option 2  is much closer. So option 1 will be correct.

Posted by

Ritika Jonwal

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