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A coil of inductance 8.4 mH and resistance 6 Omega are connected to a 12 V battery. At what time the current in the coil will be 1.0 A ?

Answer»

SOLUTION :`i_(0)=(E)/(R) =2`, Given current, `i= (i_(0))/(2)`
`:. i=i_(0)(1-e^(-(t)/(tau))) implies (1)/(2)=1-e^(-(t)/(tau))`
`implies e^((t)/(tau))=2"":. t=tau ln 2=(L)/(R)ln2` i.e., `t=(8.4 xx 10^(-3) xx 0.6931)/(6) ~~ 1 xx 10^(-3) s`


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