A) \[0,4{{L}_{0}},2{{L}_{0}}\]
B) \[4{{L}_{0}},2{{L}_{0}},0\]
C) \[2{{L}_{0}},0,4{{L}_{0}}\]
D) \[2{{L}_{0}},4{{L}_{0}},0\]
Correct Answer: D
Solution :
When two solenoids of induttance \[{{L}_{0}}\] are connected m series at ferge distancr and curreiit i is passed through them, the total flax linkage \[{{\phi }_{total}}\] is the sum of the flux Linkages \[{{L}_{0}}i\] and \[{{L}_{0}}i\]. i.e., \[{{\phi }_{total}}={{L}_{0}}i+{{L}_{0}}i\] If L be the equivalent inductance of the system, then \[{{\phi }_{total}}=Li\] \[\therefore \] \[Li={{L}_{0}}i+{{L}_{0}}i\] or \[L=2{{L}_{0}}\] 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 emfs \[{{e}_{1}}\] and \[{{e}_{2}}\] in the respective coils, i.e., \[e={{e}_{1}}+{{e}_{2}}\] \[=\left( -{{L}_{0}}\frac{di}{dt}\pm {{L}_{0}}\frac{di}{dt} \right)\] \[+\left( -{{L}_{0}}\frac{di}{dt}\pm {{L}_{0}}\frac{di}{dt} \right)\] where \[(+)\] sign is for positive coupling and \[(-)\]sign for negative coupling. But, \[e=-L.\,\frac{di}{dt}\] \[\therefore \] \[-L\frac{di}{dt}=-{{L}_{0}}\frac{di}{dt}-{{L}_{0}}\frac{di}{dt}\pm 2{{L}_{0}}\frac{di}{dt}\] i.e, \[L={{L}_{0}}+{{L}_{0}}+2{{L}_{0}}\] \[=4{{L}_{0}}\] (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, \[L={{L}_{0}}+{{L}_{0}}-2{{L}_{0}}\] \[=0\]
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