In the circuit shown, the switch 'S' is closed at `t =0`. Then capacitor is initially uncharged. Then the current through the resistor `R = 8Omega` at `t = 0.4 xx 10^(-3) sec` is (Take `e^(-2) = 0.135` )

Options

A. `0.18A`
B. `0.27A`
C. `0.54A`
D. `0.72A`

Answer

B

Explanation

time constant , `tau = R_(eq)C_(eq)`
`= 10 xx 20 xx 10^(-6)`
`= 0.2 xx 10^(-3) sec`
The current through the battery just after closing the witch is `I_(0) = (40 xx 3)/(20) = 6A`
The current through `R = 8 Omega` register ate `t = 0` is
`i_(0) = (1)/(3) xx I_(0) = (1)/(3) xx 6 = 2A`
The current through `R = 8Omega` resistor at time 't'
`i_(0) = i_(0)e^(-t//tau)`
`i = 2 e^(-t//0.2 xx 10^(-3))`
`:. at t = 0.4 xx 10^(-3) sec.
`i = 2e^(-2) = 2 xx 0.135 = 0.27 A`