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InterviewSolution
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A current of `4A` flows in a coil when connected to a `12V DC` source. If the same coil is connected to a `12V, 50 rad//s AC` source, a current of `2.4A` flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a`2500muF` capacitor is connected in series with the coil. |
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Answer» (i) A coil consists of an inductance `(L)` and a resistance `(R)` In `DC`, only resistance is effective. Hence, `R=V/i=12/4=3Omega` In `AC`, `i_("rms")=V_("rms")/Z=V_("rms")/(sqrt(R^2+omega^2L^2))` `:.L^2=1/omega^2[(V_("rms")/i_("rms"))^2-R^2]` `:. L=1/omega[(V_("rms")/i_("rms"))^2-R^(2)` Substituting the values, we have `L=1/50sqrt((12/2.4)^2-(3)^(2))` `=0.08H` (ii) When capacitor is connected to the circuit, the impedance is `Z=sqrt(R^2+(X_L-X_C)^(2))` Here `R=3Omega` `X_L=omegaL=(50)(0.08)=4Omega` and `X_C=1/(omegaC)=1/((50)(2500xx10^-6))=8Omega` `:. Z=sqrt((3)^2+(4-8)^(2))=5 Omega` Now, `(:P:)=V_("rms")i_("rms")cosphi` `=V_("rms")xxV_("rms")/ZxxR/Z` `=(V_("rms")/Z)^2xxR` Substituting the values, we have `(:P:)=(12/5)^2xx3` `=17.28W`. |
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