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A cell of 1.5V is connected across an in...

A cell of 1.5V is connected across an inductor of 2mH in series with a 2 `Omega` resistor What is the rate of growth of current immediately after the cell is switched on.

Text Solution

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`E = L(dI)/(dt) +IR ` , Therefore, `(dI)/(dt) = (E-IR)/(L)`
E=1.5 Volt, `R=2Omega , L=2mH = 2xx10^(-3)`
When the cell is switched on, I = 0
Hence `(dI)/(dt) = (1.5)/(2 xx 10^(-3)) As^(-1) = 750 As^(-1)`
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