Question

Difficulty: HardInductors and Transformers

In an electronic filter circuit, two inductors with values L1=30 mHL_1 = 30\text{ mH} and L2=20 mHL_2 = 20\text{ mH} are connected in parallel. This parallel pair is connected in series with a third inductor L3=18 mHL_3 = 18\text{ mH} and a fourth inductor L4=10 mHL_4 = 10\text{ mH}. Assuming negligible mutual inductance between the coils, what is the total equivalent inductance of the circuit in millihenries (mH\text{mH})?

Answer: 40 mH

Answer

40 mH
The total inductance is 40 mH40\text{ mH}. Inductors follow the same combination rules as resistors: the parallel pair (30 mH30\text{ mH} and 20 mH20\text{ mH}) yields an equivalent inductance of 12 mH12\text{ mH}, which when added to the series inductors (18 mH18\text{ mH} and 10 mH10\text{ mH}) gives 12+18+10=40 mH12 + 18 + 10 = 40\text{ mH}.

Step-by-Step Solution

1
Calculate the equivalent inductance of the two parallel inductors L1L_1 and L2L_2.
Lp=L1×L2L1+L2=30 mH×20 mH30 mH+20 mH=60050=12 mHL_p = \frac{L_1 \times L_2}{L_1 + L_2} = \frac{30\text{ mH} \times 20\text{ mH}}{30\text{ mH} + 20\text{ mH}} = \frac{600}{50} = 12\text{ mH}
Inductors in parallel combine using reciprocal addition, identically to resistors in parallel.
2
Calculate the total circuit inductance by summing the parallel combination with the remaining series inductors L3L_3 and L4L_4.
Ltotal=Lp+L3+L4=12 mH+18 mH+10 mH=40 mHL_{total} = L_p + L_3 + L_4 = 12\text{ mH} + 18\text{ mH} + 10\text{ mH} = 40\text{ mH}
Inductors in series combine via direct addition because the magnetic fields and induced voltages add along the common current path.

Key Concept

Inductors combine like resistors: series inductances add directly, while parallel inductances combine using reciprocal addition.
Estimated Time:1m 30s
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