An electronics technician is assembling a tuning filter network. The circuit consists of t… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician is assembling a tuning filter network. The circuit consists of two inductors connected in series with values of $18\text{ mH}$ and $42\text{ mH}$ . This series combination is then connected in parallel with a third inductor rated at $40\text{ mH}$ . Assuming there is no mutual inductance between the coils, what is the total equivalent inductance of the circuit in millihenries ( $\text{mH}$ )?

Answer: 24 mH

Answer

The total equivalent inductance of the circuit is

24\text{ mH}

Combining the two series inductors (

18\text{ mH} + 42\text{ mH}

) yields an equivalent branch inductance of

60\text{ mH}

. Connecting this branch in parallel with the

40\text{ mH}

inductor gives a total inductance of

\frac{60 \times 40}{60 + 40} = \frac{2400}{100} = 24\text{ mH}

Step-by-Step Solution

Calculate the equivalent inductance of the series branch (

L_{\text{series}}

L_{\text{series}} = 18\text{ mH} + 42\text{ mH} = 60\text{ mH}

Inductors in series combine by direct addition, similar to series resistors (

L_{\text{series}} = L_1 + L_2

Calculate the total inductance of the series branch in parallel with the third inductor.

L_{\text{total}} = \frac{60\text{ mH} \times 40\text{ mH}}{60\text{ mH} + 40\text{ mH}} = \frac{2400}{100} = 24\text{ mH}

Inductors in parallel combine using the reciprocal or product-over-sum formula (

L_{\text{total}} = \frac{L_A \times L_B}{L_A + L_B}

Key Concept

Calculation of total inductance for series-parallel inductor combinations without mutual coupling.

Estimated Time:1m 15s

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