Changing Voltage: Current and Power

Predict what changes when only the source voltage changes.

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Use a steady ideal source and an unchanged ideal resistor. Current in amperes is I = V / R; multiply by 1,000 for milliamps. Power is P = V x I = V x V / R. At fixed resistance, multiplying voltage by a factor multiplies current by that factor and power by its square. These are paper models, not instructions for a physical circuit.

Worked example

A 300-ohm resistor at 3 V carries 3 / 300 = 0.01 A = 10 mA and dissipates 3 x 0.01 = 0.03 W. At 6 V it carries 20 mA and dissipates 0.12 W. Double voltage gives double current but four times the power, because both factors in V x I doubled.

Would you believe it?

With an unchanged ideal resistor, halving voltage halves current but quarters power. Power depends on voltage multiplied by current, not on either quantity alone.

OpenStax: electric power equations

Question 1 An ideal 200-ohm resistor has 4 V across it. What current flows? 2 mA 20 mA 800 mA

Question 2 Keep that 200-ohm resistor and raise the source to 8 V. What is the current? 40 mA 20 mA 80 mA

Question 3 Compare the same 200-ohm resistor at 4 V and 8 V. What happens to power? It doubles from 0.08 W to 0.16 W It stays at 0.08 W It rises from 0.08 W to 0.32 W: four times as much

Question 4 At 12 V across the same 200 ohms, what is the power? 0.24 W 0.72 W 0.06 W

Question 5 For an unchanged ideal resistor, halve its voltage. What happens to current and power? Current halves; power becomes one quarter Both become one half Current becomes one quarter; power halves

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