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The inductor shown in figure has inducance `0.54 H` and carries a current in the direction shown thast is decreasing at a uniform rate `(di)/(dt)=0.03A//s`. a. Find the self induced emf b. Which emf of the inductor a ro b is at a higher potential? |
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Answer» (i) Given, inductance, L= 0.54H `(di)/(dt)=-0.03` Self-induced emf is givne as `e=-L(di)/(dt)=(-0.54)(-0.03)V` `=1.62xx10^(-2)V` (ii) We know that, `V_(ba)=L(di)/(dt)=-1.62xx10^(-2)V` Since, `V_(ba)(=V_(b)-V_(a))` is negative, It implies that `V_(a)gtV_(b)` or a is at higher potential. |
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