When a choke coil carrying a steady current is short circuited- the current in it decreases to - 1 times its initial value in a time T- The time constant of the choke coil is

The correct option is D T ln( 1 β) The equation governing the current is i = i_o e^ t/ τ L nbsp;β nbsp;i_o = i_o e^ t/ τ L Take log on both si ...

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Standard XII

Physics

Transformer

When a choke ...

Question

When a choke coil carrying a steady current is short circuited, the current in it decreases to $β (< 1)$ times its initial value in a time $T$ . The time constant of the choke coil is

\frac{T}{β}

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\frac{T}{ln (\frac{1}{β})}

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\frac{T}{ln (β)}

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T ln (β)

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Solution

The correct option is D $\frac{T}{ln (\frac{1}{β})}$
The equation governing the current is
$i = i_{o} e^{- t / τ L}$
$β i_{o} = i_{o} e^{- t / τ L}$
Take log on both sides
$τ = \frac{T}{l n (1 / β)}$

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When a choke coil carrying a steady current is short circuited, the current in it decreases to β(<1) times its initial value in a time T. The time constant of the choke coil is

The correct option is D Tln(1β)The equation governing the current isi=ioe−t/τLβio=ioe−t/τLTake log on both sidesτ=Tln(1/β)

When a choke coil carrying a steady current is short circuited, the current in it decreases to $β (< 1)$ times its initial value in a time $T$ . The time constant of the choke coil is

The correct option is D $\frac{T}{ln (\frac{1}{β})}$
The equation governing the current is
$i = i_{o} e^{- t / τ L}$
$β i_{o} = i_{o} e^{- t / τ L}$
Take log on both sides
$τ = \frac{T}{l n (1 / β)}$