Question

Two ideal inductors, L1 and L2, have zero internal resistance and are far apart, so their magnetic fields do not influence each other. (a) Assuming these inductors are connected in series, show that they are equivalent to a single ideal inductor having Leq = L1 + L2. (b) Assuming these same two inductors are connected in parallel, show that they are equivalent to a single ideal inductor having 1/Leq = 1/L1 + 1/L2. (c) What If? Now consider two inductors L1 and L2 that have nonzero internal resistances R1 and R2, respectively. Assume they are still far apart, so their mutual inductance is zero, and assume they are connected in series. Show that they are equivalent to a single inductor having Leq = L1 + L2 and Req = R1 + R2. (d) If these same inductors are now connected in parallel, is it necessarily true that they are equivalent to a single ideal inductor having 1/Leq = 1/L1 + 1/L2 and 1/Req = 1/R1 + 1/R2?

Answer 1

When Two ideal inductors, L1 and L2, have zero internal resistance and are far apart, so their magnetic fields do not influence each other

(a) When two ideal inductors L1 and L2 with zero internal resistance are connected in series, their inductances add up. This is because the total magnetic flux linkage in the combined system is equal to the sum of the individual flux linkages. Mathematically, Leq = L1 + L2, so they are equivalent to a single ideal inductor with inductance Leq.

(b) When the same inductors are connected in parallel, their equivalent inductance can be found using the formula for parallel connected components: 1/Leq = 1/L1 + 1/L2. This formula shows that the reciprocal of the equivalent inductance is equal to the sum of the reciprocals of the individual inductances.

(c) For inductors L1 and L2 with nonzero internal resistances R1 and R2, when connected in series, their equivalent inductance remains Leq = L1 + L2, as mutual inductance is still zero. The equivalent resistance in series connection is the sum of individual resistances: Req = R1 + R2.

(d) When these inductors with internal resistances are connected in parallel, the formula for equivalent inductance remains the same: 1/Leq = 1/L1 + 1/L2. However, the equivalent resistance formula also follows the parallel connection rule: 1/Req = 1/R1 + 1/R2.

Therefore, it is true that these inductors are equivalent to a single inductor with 1/Leq = 1/L1 + 1/L2 and 1/Req = 1/R1 + 1/R2 when connected in parallel.