Two 470 kΩ resistors in series yield an equivalent resistance of?

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Multiple Choice

Two 470 kΩ resistors in series yield an equivalent resistance of?

In a series connection, resistances add together. So two 470 kΩ resistors in series give 470 kΩ + 470 kΩ = 940 kΩ. That matches the idea that the same current flows through both, and the total opposition to that current is the sum of the individual resistances. For contrast, a single resistor would be 470 kΩ, two in parallel would yield 235 kΩ, and four in series would give 1880 kΩ. But with two in series, the sum is 940 kΩ.

In a series connection, resistances add together. So two 470 kΩ resistors in series give 470 kΩ + 470 kΩ = 940 kΩ.

That matches the idea that the same current flows through both, and the total opposition to that current is the sum of the individual resistances. For contrast, a single resistor would be 470 kΩ, two in parallel would yield 235 kΩ, and four in series would give 1880 kΩ. But with two in series, the sum is 940 kΩ.

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