Check the dimensional correctness of V = IR
Ohm's law. The potential difference across a resistor carrying current I.
Equation checker
Matching dimensions are necessary, not sufficient. A check cannot see pure numbers such as 1/2 or 2π, so a consistent equation can still be wrong.
V = I RV[M L2 T-3 A-1]=IR[M L2 T-3 A-1]Step by step
1. What each symbol stands for
- V electric potential [M L2 T-3 A-1]
- I electric current [A]
- R electrical resistance [M L2 T-3 A-2]
2. The dimensions of each term
- V [M L2 T-3 A-1] [M L2 T-3 A-1]
- IR [A] × [M L2 T-3 A-2] [M L2 T-3 A-1]
3. Compare
Every term has dimensions [M L2 T-3 A-1], so the equation is dimensionally consistent.
What it proves and what it does not
It can show
- Every term that is added or subtracted has the same dimensions, so the sum makes sense.
- Both sides of the equals sign have the same dimensions.
- Angles inside sin, cos and similar functions are pure numbers.
It cannot show
- A pure number such as 1/2 or 2π has no dimensions, so a check cannot see a missing or wrong one.
- Two quantities with the same dimensions, such as work and torque, cannot be told apart.
- A sign error or a wrong power that keeps the dimensions equal goes unnoticed.
Matching dimensions are necessary but not sufficient. Pure numbers such as 1/2 or 2π have no dimensions, so a check cannot see them, and an equation with matching dimensions can still be wrong.
Look up each quantity
V = IR: quick answers
Is V = IR dimensionally correct?
Yes. Every term has the same dimensions, [M L² T⁻³ A⁻¹], so V = IR is dimensionally consistent. That does not prove the equation itself is correct.
Does a dimension check prove V = IR is right?
No. A check can show that an equation is wrong, but it cannot show that it is right. Pure numbers such as 1/2 and 2π have no dimensions, so the check cannot see them.