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A cell of 1.5V is connected across an inductor of 2mH in series with a 2 Omegaresistor 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=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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