When inductors are connected in series, the total inductance is

When inductors are connected in series, the total inductance is

• A. The product of the inductances
• B. The reciprocal of the sum of reciprocals
• C. The sum of the reciprocals of the inductances
• D. The sum of the individual inductances

Answer: (D)

In a series arrangement, the current flowing through each inductor is the same, so their responses to a changing current add up. Each inductor has a voltage v = L di/dt across it, and the total voltage across the string is the sum of those voltages: v_total = v1 + v2 + … = (L1 + L2 + …) di/dt. By definition, the total inductance L_eq in the series circuit relates the total voltage to the rate of change of current as v_total = L_eq di/dt. Comparing the two expressions shows that L_eq = L1 + L2 + …. The sum reflects how inductors in series collectively oppose changes in current. The other formulas correspond to inductors in parallel. In parallel, the voltages are the same across each inductor and the currents add, leading to L_eq = 1 / (1/L1 + 1/L2 + …). The product of inductances isn’t the correct way to combine them in series, reinforcing why the sum is the right result here.

In a series arrangement, the current flowing through each inductor is the same, so their responses to a changing current add up. Each inductor has a voltage v = L di/dt across it, and the total voltage across the string is the sum of those voltages: v_total = v1 + v2 + … = (L1 + L2 + …) di/dt. By definition, the total inductance L_eq in the series circuit relates the total voltage to the rate of change of current as v_total = L_eq di/dt. Comparing the two expressions shows that L_eq = L1 + L2 + …. The sum reflects how inductors in series collectively oppose changes in current.

The other formulas correspond to inductors in parallel. In parallel, the voltages are the same across each inductor and the currents add, leading to L_eq = 1 / (1/L1 + 1/L2 + …). The product of inductances isn’t the correct way to combine them in series, reinforcing why the sum is the right result here.