Physics - EMI/AC Question with Solution | TestHub

When two solenoids of inductance$L_0$are connected in series at large distance and current$i$is passed through them, the total flux linkage${\mathrm{\varphi }}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}$is the sum of the flux linkages$L_{0}i$and$L_{0}i, ie,$

$\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{\varphi }}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=L_{0}i+L_{0}i$

If$L$be the equivalent inductance of the system, then

$\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{\varphi }}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=Li$

$\therefore Li=L_{0}i+L_{0}i$

or$L=2L_0$

When solenoids are connected in series with one inside the other and senses of the turns coinciding, then there will be mutual inductance$L$between them. In this case the resultant induced emf in the coils is the sum of the emf$\mathrm{s}\mathrm{ }{e}_1$and$e_2$in the respective coils,$ie,$

$e=e_1+e_2$

$=\left(-L_0\frac{di}{dt}\pm L_0\frac{di}{dt}\right)+\left(-L_0\frac{di}{dt}\pm L_0\frac{di}{dt}\right)$

Where (+) sign is for positive coupling and (-) sign for negative coupling.

But,$e=-L.\frac{di}{dt}$

$\therefore -L\frac{di}{dt}=-L_0\frac{di}{dt}-L_0\frac{di}{dt}\pm 2L_0\frac{di}{dt}$

$ie, L=L_0+L_0+2L_0$

$=4L_0$(for positive coupling)

When solenoids are connected in series with one inside the other with senses of the turns opposite, then their is negative coupling.

So,$L=L_0+L_2-2L_0=0$