Question

A coil of inductance 0.2 henry is connected to 600 Volt battery .At What rate ,will the Current in the coil grow when circuit is completed ??​

Answer 1

Explanation:

$\tt Inductance \: (L) = 0.2\: Henry \\ \tt Voltage\:(e) = 600 \:V \\ \\ \tt We\:know\:that \\ \\ \tt |e| = L\frac{dI}{dt} \\ \\ \tt Rearranging\:the\:above\: equation\:to\:find\:rate\:of\:current \\ \\ \tt \frac{dI}{dx} = \frac{|e|}{L} \\ \\ \tt \longrightarrow \frac{dI}{dt} = \frac{600}{0.2} \\ \\ \tt \longrightarrow \frac{dI}{dt} = 3000 \:A\:{s}^{-1} \\ \\ \boxed{\bold{\underline{\red{\tt At \: 3000\:A\:{s}^{-1}\:rate\:the\: current\:in\:coil\:grow}}}}$

Answer 2

Explanation:

Inductance(L)=0.2Henry

Voltage(e)=600V

Weknowthat

∣e∣=L

dt

dI

Rearrangingtheaboveequationtofindrateofcurrent

dx

dI

=

L

∣e∣

⟶

dt

dI

=

0.2

600

⟶

dt

dI

=3000As

−1

At3000As

−1

ratethecurrentincoilgrow