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Find the equivalent inductance of two inductors having inductances L_1 and L_2 connected in parallel with the help of appropriate DC circuit. |
Answer» Solution :Let two inductor `L_1` and `L_2` are connected in parallel across a battery whose emf can be varied continuously and having no INTERNAL resistance.  Let currents passing through `L_1` and `L_2` are `I_1` and `I_2` at time t and the rates of change of currents through them are `(dI_1)/(DT)` and `(dI_2)/(dt)` respectively. `therefore` p.d. between their two ENDS are respectively `-L_1 (dI_1)/(dt)` and `-L_2(dI_2)/(dt)` Current passing through the main circuit is I at time t then, `epsilon =-L (dI)/(dt)=-L ((dI_1)/(dt)+(dI_2)/(dt))`...(1) But, `L_1(dI_1)/(dt)=-epsilon` and `L_2(dI_2)/(dt)=-epsilon` `therefore (dI_1)/(dt)=-epsilon/L_1` and `(dI_2)/(dt)=-epsilon/L_2` Putting these values in eqn. (i) `epsilon=-L (-epsilon/L_1-epsilon/L_2)` `therefore 1/L=1/L_1+1/L_2` |
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