The current in an L - R circuit- in a time t - 2LR reduces toTake e - 2-731

In L R circuit, current as a function of time (t) is given by i = i_0 e^ Rt/L=i_0e^ RL×2LR At t = 2LR i=i_0e^ 2 = i_0e^2 = 0.134i_0 = 13.4%. ...

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Standard XII

Physics

Simple Circuit

The current i...

Question

The current in an $L - R$ circuit, in a time $t = \frac{2 L}{R}$ reduces to
(Take $e = 2.731$ )

36.5 %

of maximum

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13.4 %

of maximum

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0.50 %

of maximum

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63.2 %

of maximum

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Solution

The correct option is B $13.4 %$ of maximum
In $L - R$ circuit,
current as a function of time $(t)$ is given by
$i = i_{0} e^{- R t / L} = i_{0} e^{- \frac{R}{L} \times \frac{2 L}{R}}$
At $t = \frac{2 L}{R}$
$i = i_{0} e^{- 2} = \frac{i_{0}}{e^{2}} = 0.134 i_{0} = 13.4 %$ .

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The current in an L−R circuit, in a time t=2LR reduces to (Take e=2.731)

The correct option is B 13.4% of maximumIn L−R circuit, current as a function of time (t) is given by i=i0 e−Rt/L=i0e−RL×2LR At t=2LR i=i0e−2=i0e2=0.134i0=13.4%.

The current in an $L - R$ circuit, in a time $t = \frac{2 L}{R}$ reduces to
(Take $e = 2.731$ )

The correct option is B $13.4 %$ of maximum
In $L - R$ circuit,
current as a function of time $(t)$ is given by
$i = i_{0} e^{- R t / L} = i_{0} e^{- \frac{R}{L} \times \frac{2 L}{R}}$
At $t = \frac{2 L}{R}$
$i = i_{0} e^{- 2} = \frac{i_{0}}{e^{2}} = 0.134 i_{0} = 13.4 %$ .