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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?

Answer»


SOLUTION :Equivalent self-inductance of two inductors connected in parallel can be written as follows:
`(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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