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Difficulty: EasyParallel Circuits

Two resistors with values of 12 Ω12\ \Omega and 6 Ω6\ \Omega are wired in parallel across a circuit board. What is the total equivalent resistance of this combination?

  1. A
    18 Ω18\ \Omega
  2. 4 Ω4\ \OmegaAnswer
  3. C
    9 Ω9\ \Omega
  4. D
    72 Ω72\ \Omega

Answer

4 Ω4\ \Omega
For two resistors in parallel, the equivalent resistance is calculated by dividing their product by their sum: 12×612+6=7218=4 Ω\frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega. In parallel circuits, the equivalent resistance is always lower than any single branch resistance.

Step-by-Step Solution

1
Identify the given branch resistances and select the parallel equivalent resistance formula.
R1=12 ΩR_1 = 12\ \Omega, R2=6 ΩR_2 = 6\ \Omega, and Req=R1×R2R1+R2R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}
The product-over-sum formula determines the total equivalent resistance of two parallel branches.
2
Calculate the product and sum of the two resistors, then divide.
Req=12×612+6=7218=4 ΩR_{eq} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega
Completing the arithmetic yields the equivalent resistance, which must be less than the smallest individual branch.

Key Concept

Total equivalent resistance in a parallel circuit is always less than the lowest individual branch resistance and can be calculated using the reciprocal or product-over-sum formula.
Estimated Time:45s
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Topics

Parallel Circuits

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