An electronics technician connects three inductors in parallel across an AC filter circuit… — Armed Services Vocational Aptitude Battery (ASVAB)

An electronics technician connects three inductors in parallel across an AC filter circuit. The inductors have inductance values of $60\text{ mH}$ , $30\text{ mH}$ , and $20\text{ mH}$ , with negligible mutual inductance between them. What is the total equivalent inductance of the circuit in millihenries (mH)?

Answer: 10 mH

Answer

The total equivalent inductance of the parallel circuit is

10\text{ mH}

Inductors in parallel without mutual coupling combine according to the reciprocal formula

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

. Computing

\frac{1}{60} + \frac{1}{30} + \frac{1}{20} = \frac{6}{60} = \frac{1}{10}

, and inverting yields

10\text{ mH}

Step-by-Step Solution

State the reciprocal formula for inductors connected in parallel

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

When inductors are in parallel and uncoupled, the total rate of current change is the sum of the individual branch rates, resulting in a reciprocal combination rule identical to parallel resistors.

Substitute the individual inductance values

\frac{1}{L_{\text{total}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}

Direct substitution of the known branch values:

L_1 = 60\text{ mH}

L_2 = 30\text{ mH}

, and

L_3 = 20\text{ mH}

Find the common denominator and sum the terms

\frac{1}{L_{\text{total}}} = \frac{1 + 2 + 3}{60} = \frac{6}{60} = \frac{1}{10}\text{ mH}^{-1}

Using the lowest common denominator (

60

) to perform fractional addition.

Invert the reciprocal to obtain equivalent inductance

L_{\text{total}} = 10\text{ mH}

Taking the reciprocal of

\frac{1}{10}\text{ mH}^{-1}

gives

10\text{ mH}

Key Concept

Parallel Inductance Calculation

Estimated Time:1m 0s

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