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Grade: 12th pass
        
An ideal coil of 10 H is connected in a series with a resistance of 5 Ω and bettery of 5 V. Two second after the connection is made, the current flowing in amperes in the circuit is
(a) e    (b) e^-1    © (1- e^-1)     (d) (1-e)
one year ago

Answers : (1)

jakira
43 Points
							
I=I0[1eRt/L]
I=I0I0[1-e-Rt/L]
                                   I0 =     5/5=1
               So                I   = 1-e-5t/10
          i.e                       I  = 1-e-t/2
As t=2,                   So   I =1-e-1
 
 
I=I0[1eRt/L]

 

one year ago
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