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Calculate the resultant inductance of two inductor L_(1) and L_(2) when they are connected in parallel in A.C. circuit. |
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Answer» Solution :Let `Z_(L_(1))` and `Z_(L_(2))` be the inductive reactance of two coils. SINCE they are connected in parallel,the resultant reactance will be, `Z= ( Z_(L_(1)) Z_(L_(2)))/( L_(L_(1))+ Z_(L_(2)))= ( J X_(L_(1)) xx jX_(L_(2)))/(jX_(L_(1))+jX_(L_(2)))` `:. Z = (jX_(L_(1))X_(L_(2)))/(X_(L_(1))+ X_(L_(2)))`.....(i) If the resultant inductance is equal to L, then `Z = j OMEGA L`. `X_(L _(1) )= omegaL_(1)` and `X_(L_(2)) = omega L_(2)` then Putting this VALUES in equation (1). `:. j omega L = ( j omega^(2) L_(1) L_(2))/( omega L_(1) + omegaL_(2))` `:. L = ( L_(1) L_(2))/( L_(1) + L_(2))` |
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