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An inductor-resistance -battery circuit ...

An inductor-resistance -battery circuit is switched on at t = 0. If the emf of the battery in one time constant `tau`.

A

`(E)/(R)*(tau)/(e)`

B

`(E)/(R)*(tau)/(e-1)`

C

`(E)/(R)*(tau)/(e+1)`

D

`(E)/(R)*(e-1)/(tau)`

Text Solution

Verified by Experts

The correct Answer is:
A
Current at time t is given by
`i=i_(0)(1-e^(-t//(tau)))where (i_0)=E//R`
The charge passed through the bettery during the period t to t+dt is idt. The total charge passed during time 0 to `tau` is
`Q=int_(0)^(tau) idt = i_(0)i int_(0)^(tau) (1-e^(-t//tau))dt`
`=i_(0)[t+(e^(-t//tau))/(tau)]_(0)^(tau) =i_(0)`
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