Two inductors L1 and L2 are connected in parallel to a battery. These two induct

Question

Two inductors L1 and L2 are connected in parallel to a battery. These two inductors act as one equivalent inductance Leq. To find Leq we first notice that because they are connected in parallel, the voltage across the two must be the same, but the rate at which the current changes with time ?I/?t is different for the two inductors.

Use these facts to write the total voltage across the two inductors in the form
V =-Leq(?I/?t) to find Leq in terms of L1 and L2. (Do not fret about minus signs too much.)

Explanation / Answer

Given Inductors L1 and L2 are connected in parallel combination . Total voltage across the two inductors is                V = L eq ( dI / dt )          ...... (1) Rate at which the current changes with time in inductance L1 is            dI1 / dt = V / L1 Rate at which the current changes with time in inductance L2 is             dI1 / dt = V / L2 Total rate of current               dI / dt = dI1 / dt + dI2 /d t                          = V / L1 + V / L2                          = V ( L 1 + L 2 ) / L1 L2                      V = ( L1 L2 / L1+ L2 ) (dI / dt ) compare to equation (1)                      Leq = L1 L2 / L1 + L2 Rate at which the current changes with time in inductance L2 is             dI1 / dt = V / L2 Total rate of current               dI / dt = dI1 / dt + dI2 /d t                          = V / L1 + V / L2                          = V ( L 1 + L 2 ) / L1 L2                      V = ( L1 L2 / L1+ L2 ) (dI / dt ) compare to equation (1)                      Leq = L1 L2 / L1 + L2

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