A current of 6 A flows in a wire. Which statement says what that means?
A wall outlet supplies 120 V between its two contacts. What does the 120 state?
A camping lantern's bulb carries 0.8 A with 6 V across it. You can now calculate its power from P = IV. But why does multiplying a current by a voltage give a rate of energy transfer? The lantern shows the whole story.
The lantern: 0.8 A through the bulb, 6 V across it.
A current of 0.8 A means 0.8 C of charge passes through the bulb every second.
A voltage of 6 V across the bulb means every coulomb transfers 6 J as it passes through.
Each second: 0.8 C passes through the bulb, and every coulomb of it transfers 6 J.
So each second, 0.8 C passes, and each coulomb of it transfers 6 J. The energy transferred each second is 0.8 × 6 = 4.8 J.
Coulombs per second times joules per coulomb gives joules per second.
Energy transferred each second is power. The bulb's power is 4.8 W, exactly the current multiplied by the voltage.
That is why P = IV works: current is the charge passing per second, and voltage is the energy transferred per coulomb, so their product is the energy transferred per second.
The units tell the same story: A × V = C/s × J/C = J/s = W. The coulombs cancel, leaving joules per second.
Each second, 0.4 C passes through a buzzer, and each coulomb transfers 9 J. Enter the energy the buzzer receives each second, in joules.
Why does multiplying the current by the voltage give the energy a component receives each second?
Two bulbs each have 6 V across them. Bulb A carries twice the current of bulb B. How fast does bulb A receive energy, compared with bulb B?
Your turn
Each second, 2.5 C passes through a heater coil, and each coulomb transfers 20 J. Calculate the rate at which the coil receives energy, in J/s.
In P = IV, what does the current I tell you in the energy story?