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The time constant of a certain inductive coil was found to be 2.5 ms. With a resistance of 80Omegaadded 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 xxx 10^(-3) sec to (1)` For th second case , `(L)/(R+80) = 0.5 xx 10^(-3) to (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 xx 10^(-3) xx 20 = 50mH` |
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