Consider the circuit shown below. When switch `S_(1)` is closed, let I be the current at time t, then applying Kirchhoff's law
`E-iR-L (di)/(dt) = 0 or int_(0)^(i) (di)/(E-iR) = 1/L int_(0)^(1) dt`
`i=E/R (1-e^(-R/l *t))`
`L/R` = time constant of circuit
When current reaches its steady value `(=i_(0)`, open `S_(1)` and close `S_(2)`, the current does reach to zero finally but decays expnentially. The decay equation is given as `i=i_(0)e^(-R/L *T)`.

Three idenctical rings. The first (a). slips without rolling and the second (b) rolls without slipping and the third rolls with slipping .

Options

A. The same emf is induced in all three rings.
B. No emf is induced in any of the rings.
C. In each ring all points are at same potential
D. B develops max. Induced emf and A, the least.

Answer

A

Explanation

Current in L-R circuit is given as `i=i_(0)(1-e^(-t//tau))`
where `i_(0)=E/R and tau =L/R`
`Q=int_(0)^(tau) (1-e^(-t//tau)) dt =i_(0)[t-(e^(-t//tau))/(-1//tau)]_(0)^(tau) = (i_(0)tau)/(e)`
`Q = i_(0) tau//e`.