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  • There are two solenoids of same length and inductance L but their diameters differ to the extent that one can just fit into the other. They are connected in three different ways in series:<b

There are two solenoids of same length and inductance L but their diameters differ to the extent that one can
just fit into the other. They are connected in three different ways in series:
(a) They are connected in series but separated by large distance
(b) They are connected in series with one inside the other and senses of the turns coinciding
(c) Both are connected in series with one inside the other with senses of the turns opposite
as depicted in figures I, II and III, respectively. The total inductance of the solenoids in each of the case I, II
and III are respectively.

 

Option: 1

0,4L0,2L0


Option: 2

4L0,2L0,0


Option: 3

2L0,0,4L0

 


Option: 4

2L0,4L0,0


Answers (1)

best_answer

When two solenoids of inductance \mathrm{L_{0}} are connected in series at large distance and current i is passed through them, then the total flux linkage \mathrm{\phi_{\text {weal }}=\mathrm{L}_0 \mathrm{i}+\mathrm{L}_0 \mathrm{i}}

If L be the equivalent inductance of the system, then

\mathrm{\begin{aligned} & \phi_{\text {locel }}=\mathrm{Li} \\ & \therefore \mathrm{Li}=\mathrm{L}_0 \mathrm{i}+\mathrm{L}_0 \mathrm{i} \quad \text { or } \quad \mathrm{L}=2 \mathrm{~L}_0 \end{aligned}}

When solenoids are connected in series with one inside the other and senses of the turns coinciding, then there will be a mutual inductance L between them. In this case, the resultant induced emf in the coils is the sum of the emf’s \mathrm{e_{1}} and \mathrm{e_{2}} in the respectively coils, i.e.,

\mathrm{\begin{aligned} & \mathrm{e}=\mathrm{e}_1+\mathrm{e}_2 \\ & =\left(-\mathrm{L}_0 \frac{\mathrm{di}}{\mathrm{dt}} \pm \mathrm{L}_0 \frac{\mathrm{di}}{\mathrm{dt}}\right)+\left(-\mathrm{L}_0 \frac{\mathrm{di}}{\mathrm{dt}} \pm \mathrm{L}_0 \frac{\mathrm{di}}{\mathrm{dt}}\right) \end{aligned}}

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

\mathrm{\begin{aligned} & e=-L \frac{d i}{d t} \\ &\therefore -L \frac{d i}{d t}=-L_0 \frac{d i}{d t}-L_0 \frac{d i}{d t} \pm 2 L_0 \frac{\mathrm{di}}{\mathrm{dt}} \end{aligned}}

But , \mathrm{\mathrm{L}=\mathrm{L}_0+\mathrm{L}_0+2 \mathrm{~L}_0=4 \mathrm{~L}_0}            [for positive coupling]

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

so, \mathrm{\mathrm{L}=\mathrm{L}_0+\mathrm{L}_0-2 \mathrm{~L}_0=0}

 

Posted by

seema garhwal

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