Given `L_(1) = 1 mH`, `R_(1) = 1 Omega`, `L_(2) = 2 mH`, `R_(2) = 2 Omega`


Neglecting mutual inductance, the time constants (in ms) for circuits (i), (ii), and (iii) are

Options

A. `1,1,(9)/(2)`
B. `(9)/(4),1,1`
C. `1,1,1`
D. `1,(9)/(4),1`

Answer

A

Explanation

Given, `L_(1) = 1 mH, L_(2) = 2 mH, R_(1) = 1 Omega, R_(2) = 2 Omega`
In the first circuit,
`L = L_(1) + L_(2)` and `R = R_(1) + R_(2)`
`tau = (L)/(R ) = (3 mH)/(3 Omega) = 1 ms`
In the second circuit,
`L = (L_(1)L_(2))/(L_(1) + L_(2)) = (2 xx 10^(-6))/(3 xx 10^(-3)) = (2)/(3) xx 10^(-3)`
`R = (R_(1)R_(2))/(R_(1) + R_(2)) = (2)/(3) Omega`
`tau_(2) = ((2)/(3) xx 10^(-3))/(2//3) = 1ms`
In the third circuit,
`L = L_(1) + L_(2) = 3 mH`
`R = (R_(1)R_(2))/(R_(1) + R_(2)) = (2)/(3) Omega`
`tau_(3) = (L)/(R ) = (3 xx 10^(-3))/(2//3) = (9)/(2) ms`