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The time constant of a certain inductive coil was found to be 2.5 ms. With a resistance of 80Omega added in series, a new time constant of 0.5 ms was obtained. Find the inductance and resistance of the coil. |
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Answer» Solution :Time CONSTANT, `tau_(L)= (L)/(R)` For the first case, `(L)/(R)= 2.5 XX 10^(-3) sec to (1)` For the second case, `(L)/(R+80)=0.5xx10^(-3) rarr (2)` Divide (1) by (2) `(R +80)/(R) =(2.5)/(0.5)=5` SOLVING we get: `R=20 Omega` Now, `(L)/(R) = 2.5 xx 10^(-3)` `L=2.5 xx10^(-3)xx20 = 50`mH |
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