In the circuit shows in Fig `E = 15 V`, `R_(1) = 1Omega`, `R_(2) = 1 Omega`, `R_(3) = 2Omega`, and `L = 1.5 H`. The currents flowing through `R_(1),R_(2)`,and `R_(3)` are `i_(1),i_(2)`, and `i_(3)`, respectively.

Immediately after tuning switch `S` on,

Options

A. `i_(1) = i_(2) = 7.5A`, `i_(3) = 0A`
B. `i_(1) = i_(3) = 5A`, `i_(2) = 0A`
C. `i_(1) = i_(2) = 9A`, `i_(3) = 0A`
D. `i_(1) = i_(2) = i_(3) = 0A`

Answer

A

Explanation

At `t = 0`, circuit can be considered as follows:

After a long, circuit can be considered a sfollow:

Immediately after closing `S`, `(di_(3))/(dt) != 0`. Because induced e.m.f.
`(Ldi_(3))/(dt) != 0`.