When inductor is short circuited then current drops to 1/n times the maximum value. The time constant of the circuit is

Question

When inductor is short circuited then current drops to 1/n times the maximum value. The time constant of the circuit is (A) τ=(t/ln n) (B) τ=(ln n/t) (C) τ=t ln n (D) τ=ln n

Tardigrade Admin · Accepted Answer

During decay, the value of current at any time

t

in

R - L

circuit is given by

I = I_{0^{e^{- t / τ}}}

where

τ

is the time constant of

R - L

circuit and

I_{0}

is the maximum current

∴ \frac{I _{0}}{n} = I_{0^{e^{- t τ}}} \dots (i)

Taking natural logarithms on both sides of eqn

(i)

, we get
ln

n = \frac{t}{τ}

or

τ = \frac{t}{l n n}

Q. When inductor is short circuited then current drops to $1/ n$ times the maximum value. The time constant of the circuit is

$τ = \frac{t}{l n n}$

$τ = \frac{l n n}{t}$

$τ = t l n n$

$τ = l n n$

Q. When inductor is short circuited then current drops to 1/n times the maximum value. The time constant of the circuit is

τ=lnnt​

τ=tlnn​

τ=tlnn

τ=lnn

During decay, the value of current at any time t in R−L circuit is given by I=I0e−t/τ​ where τ is the time constant of R−L circuit and I0​ is the maximum current ∴nI0​​=I0e−tτ​…(i) Taking natural logarithms on both sides of eqn (i), we get ln n=τt​ or τ=lnnt​

Q. When inductor is short circuited then current drops to $1/ n$ times the maximum value. The time constant of the circuit is

$τ = \frac{t}{l n n}$

$τ = \frac{l n n}{t}$

$τ = t l n n$

$τ = l n n$