Question

A coil of inductance 8.4 milli henry and resistance 6 ohm is connected to a 12 volt battery in series the current in the coil is 1:00 ampere at approximately the time

Answer 1

Time taken will be equal to $t=9.7\times 10^{-4}sec$

Explanation:

We have given inductance of the coil $L=8.4mH=8.4\times 10^{-3}H=0.0084H$

Resistance R = 6 ohm

So time constant $\tau =\frac{L}{R}=\frac{0.0084}{6}=0.0014sec$

Emf of the battery V = 12 volt

Current in the circuit i = 1 amp

Current in the RL circuit is equal to $i=\frac{V}{R}(1-e^{\frac{-t}{\tau }})$

So $1=\frac{12}{6}(1-e^{\frac{-t}{0.0014 }})$

$0.5=(1-e^{\frac{-t}{0.0014 }})$

$e^{\frac{-t}{0.0014 }}=0.5$

${\frac{-t}{0.0014 }}=ln0.5$

${\frac{-t}{0.0014 }}=-0.6931$

$t=9.7\times 10^{-4}sec$

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