In a parallel circuit, the total resistance is less than that of the lowest value resistor.

Study for the MindTap AC/DC Test. Use flashcards and multiple-choice questions, each offering hints and explanations. Prepare to ace your exam!

Multiple Choice

In a parallel circuit, the total resistance is less than that of the lowest value resistor.

Explanation:
In parallel circuits, having multiple paths for current lowers the overall resistance. The total resistance is found by adding the reciprocals of each resistor and then taking the reciprocal of that sum: 1/R_total = 1/R1 + 1/R2 + … . Because you’re adding positive numbers, the sum is larger than any single 1/Ri, which makes R_total smaller than the smallest individual resistor. That’s exactly what the statement describes, so the quantity in question is resistance. The other concepts don’t describe this behavior: voltage across parallel branches is the same for each branch, not the total resistance; capacitance relates to capacitors (not the resistors here); current is the total current that splits among the branches, not the total resistance. For a quick check, two resistors in parallel, 6 ohms and 3 ohms, give 1/R_total = 1/6 + 1/3 = 1/2, so R_total = 2 ohms, which is indeed less than the smallest resistor.

In parallel circuits, having multiple paths for current lowers the overall resistance. The total resistance is found by adding the reciprocals of each resistor and then taking the reciprocal of that sum: 1/R_total = 1/R1 + 1/R2 + … . Because you’re adding positive numbers, the sum is larger than any single 1/Ri, which makes R_total smaller than the smallest individual resistor. That’s exactly what the statement describes, so the quantity in question is resistance.

The other concepts don’t describe this behavior: voltage across parallel branches is the same for each branch, not the total resistance; capacitance relates to capacitors (not the resistors here); current is the total current that splits among the branches, not the total resistance. For a quick check, two resistors in parallel, 6 ohms and 3 ohms, give 1/R_total = 1/6 + 1/3 = 1/2, so R_total = 2 ohms, which is indeed less than the smallest resistor.

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