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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. |
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Answer» Solution :`E=L(dI)/(dt)+IR`, THEREFORE, `(dI)/(dt)=(E-IR)/(L)` E=1.5 Volt,R=2 `OMEGA`, L = 2mH = `2 xx 10^(-3)` H When the cell is SWITCHED on, I = 0 Hence `(dI)/(dt)=(E)/(L)=(1.5)/(2 xx 10^(-3)) As^(-1)= 750 As^(-1)` |
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