Questions
Two pure inductors, each of self-inductance L are connected in parallel but are well separated from each other, then the total inductance is
detailed solution
Correct option is C
short cut: 1Leff=1L1+1L2When the coils are connected in parallel, let the currents in the two coils be i1 and i2 respectively. Total induced currenti=i1+i2 or didt=di1dt+di2dt ........(1) Now e=−L1di1dt=−L2di2dt[∵ In parallel, induced e.m.f across each coil will be same) Hence di1dt=−eL1 and di2dt=−eL2 .........(2)Let L' be the equivalent inductance. Thene=−L′didt or didt=−eL′ ........(3)From eqs. (1), (2) and (3), we get−eL′=−eL1−eL2 or 1L′=1L1+1L2 ∴ L′=L1I2L1+L2 Here L1=L2=L∴ L′=L×LL+L=L2Talk to our academic expert!
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