Two pure inductors, each of self-inductance L are connected in parallel but are well separated from each other, then the total inductance is

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=L2