WEBVTT

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Here is the problem: Resistors of 5, 5, 5, 5 Ω are connected in series to a 10 V battery. Find the total resistance and current.

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Resistors in series carry the same current; their resistances add.

3
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Series: resistances add.

4
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Total current from Ohm prime s law.

5
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Voltage across each resistor (same current). The drops add up to the supply voltage.

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Power delivered.

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So the answer is R T equals 20 Ω; I equals 0.5 A, P equals 5 W.

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The equivalent resistance of a series chain is always larger than its largest resistor.

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R T greater than or equal to every individual resistor, which checks out.