A coil of inductance 0.5 Henry is connected to 18 V battery .calculate the rate of growth of current

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Solution

$Given, Self inductance of coil, L = 0.5 Henry Voltage of the battery, V = 18 V From the relation V = L \frac{d I}{d t} \Rightarrow 18 = 0.5 \frac{d I}{d t} \Rightarrow \frac{d I}{d t} = 36 A - s$

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A coil of inductance 0.5 Henry is connected to 18 V battery .calculate the rate of growth of current

Given,Self inductance of coil, L=0.5 HenryVoltage of the battery, V =18 VFrom the relationV=LdIdt⇒18=0.5dIdt⇒dIdt=36 A-s

$Given, Self inductance of coil, L = 0.5 Henry Voltage of the battery, V = 18 V From the relation V = L \frac{d I}{d t} \Rightarrow 18 = 0.5 \frac{d I}{d t} \Rightarrow \frac{d I}{d t} = 36 A - s$