A 60 Ω lamp carries 0.20 A. Enter the voltage across it in V.
A heated car seat's element carries 3.0 A, and its resistance is 4.0 Ω. The voltage across the element is not given, and P = IV needs it. A second substitution into P = IV removes the need.
The meter gives the current and the label gives the resistance; the voltage across the element is not marked.
Here is the two-step route: find the voltage with V = IR, then use P = IV.
The two-step route: V = IR finds the voltage, then P = IV finds the power.
The two steps land on 36 W. Now run the same two steps with symbols instead of numbers.
The voltage across the element is V = IR.
Substitute IR for V in P = IV: P = I × IR.
I × I is I². So P = I²R.
Both routes give 36 W; P = I²R does it in one substitution.
P = I²R gives the power dissipated, the power the element transfers away as thermal energy, straight from the current through it and its resistance.
Substituting V = IR into P = IV gives...
The current appears twice in P = I²R: the current carries the charge each second, and the current also sets the voltage IR across the element.
So doubling the current through a fixed resistance quadruples the power dissipated.
The current through a fixed resistor triples. The power dissipated in it becomes...
The element heats because current passing through resistance transfers electrical energy to thermal energy, and the heating rises sharply as the current rises.
In P = I²R, which quantity is squared?
Your turn
The current through a motor winding halves while its resistance stays the same. The power dissipated in the winding becomes...