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Two inductors are connected in parallel to each other and their resultant is found to be 2.4 H. If one of themis having self-inductance of 6 H then what will be selfinductance of the other inductor in henry? |
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Answer» `(L_(1)L_(2))/(L_(1)+L_(2))` `(L_(1)L_(2))/(L_(1)+L_(2))= 2.4` `L_(1)= 6H` `IMPLIES (6L_(2))/(6+L_(2))= 2.4 implies L_(2)= 2.4 +0.4L_(2)implies 0.6 xx L_(2)=2.4` `implies L_(2)=4` |
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