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  • A coil of inductance 8.4 mH and resistance 6 is connected to a 12 V battery. The current in the coil is one.

A coil of inductance 8.4 mH and resistance 6\Omega is connected to a 12 V battery. The current in the coil is one. A at approximately the time :

Option: 1

500 s


Option: 2

20 s


Option: 3

35 ms


Option: 4

1 ms


Answers (1)

best_answer

The current-time (i–t) equation in L-R circuit is given by [Growth of current in L-R circuit]
\mathrm{\begin{aligned} & \mathrm{I}=\mathrm{i}_0\left[1-\mathrm{e}^{-\mathrm{t} / \mathrm{L} L}\right] \\ & \text { where } \quad \mathrm{i}_0=\frac{V}{R}=\frac{12}{6}=2 \mathrm{~A} \\ & \text { and } \quad \tau_L=\frac{L}{R}=\frac{8.4 \times 10^{-3}}{6} \\ & =1.4 \times 10^{-3} \mathrm{~s} \\ & \text { and } \\ & \mathrm{i}=1 \mathrm{~A} \\ & t=\text { ? } \\ & \end{aligned} }
Substituting these values in Eq. (1), we get
\mathrm{ \text { or } \quad \begin{aligned} & \mathrm{t}=0.97 \times 10^{-3} \mathrm{~s} \\ & \mathrm{t}=0.97 \mathrm{~ms} \\ & \mathrm{t} \approx 1 \mathrm{~ms} \end{aligned} }

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