FCC Exam Question: 6A68
What is the total resistance of a parallel circuit consisting of a 10 ohm branch and a 25 ohm branch?
Explanation: When resistors are connected in parallel, the total resistance of the circuit decreases. For two resistors, $R_1$ and $R_2$, in parallel, the equivalent resistance ($R_{total}$) is calculated using the product-over-sum formula: $R_{total} = (R_1 \times R_2) / (R_1 + R_2)$ Plugging in the given values: $R_{total} = (10 \, \Omega \times 25 \, \Omega) / (10 \, \Omega + 25 \, \Omega)$ $R_{total} = 250 \, \Omega^2 / 35 \, \Omega$ $R_{total} \approx 7.1428 \, \Omega$ Rounded to two decimal places, this is 7.14 ohms, which corresponds to option D. This fundamental principle of parallel circuits ensures that adding more branches provides additional paths for current, thus reducing the overall opposition to current flow. Option B is incorrect because the total resistance in a parallel circuit must always be less than the smallest individual branch resistance (10 ohms). Option C is a rounding that is not as precise as D.
6A593
6A78
6A125
6A212
6A504
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Includes Elements 1, 3, 6, 7R, 8, and 9.