Problem
Convert milliamps to amps and state a general formula for converting any current value written in mA into amperes (A).
Key idea: SI prefix “milli-”
The SI prefix milli- means “one thousandth”: \[ 1\ \text{mA} = 10^{-3}\ \text{A} \] Equivalently, \[ 1\ \text{A} = 10^{3}\ \text{mA} \]
General conversion formula (mA to A)
Start from \(1\ \text{mA} = 10^{-3}\ \text{A}\). Multiply a current \(I(\text{mA})\) by the factor \(\frac{10^{-3}\ \text{A}}{1\ \text{mA}}\) so that mA cancels:
\[ I(\text{A}) = I(\text{mA})\cdot \frac{10^{-3}\ \text{A}}{1\ \text{mA}} = I(\text{mA})\cdot 10^{-3} = \frac{I(\text{mA})}{1000} \]
Step-by-step examples
Example 1: Convert \(250\ \text{mA}\) to amps.
\[ 250\ \text{mA}\cdot 10^{-3}\ \frac{\text{A}}{\text{mA}} = 250\cdot 10^{-3}\ \text{A} = 0.250\ \text{A} \]
Example 2: Convert \(12\ \text{mA}\) to amps.
\[ 12\ \text{mA} = \frac{12}{1000}\ \text{A} = 0.012\ \text{A} \]
Why this matters in physics circuits
In circuit analysis, current is typically substituted into formulas in amperes. For example, in Ohm’s law, \[ V = I\cdot R \] the standard SI convention is \(I\) in amperes, \(R\) in ohms, and \(V\) in volts. A value reported in milliamperes must be converted to amps before direct substitution to keep units consistent.
Quick reference table
| Current (mA) | Conversion | Current (A) |
|---|---|---|
| \(1\ \text{mA}\) | \(1\cdot 10^{-3}\ \text{A}\) | \(0.001\ \text{A}\) |
| \(10\ \text{mA}\) | \(10\cdot 10^{-3}\ \text{A}\) | \(0.010\ \text{A}\) |
| \(100\ \text{mA}\) | \(100\cdot 10^{-3}\ \text{A}\) | \(0.100\ \text{A}\) |
| \(500\ \text{mA}\) | \(500\cdot 10^{-3}\ \text{A}\) | \(0.500\ \text{A}\) |
| \(1000\ \text{mA}\) | \(1000\cdot 10^{-3}\ \text{A}\) | \(1.000\ \text{A}\) |
Visualization: SI prefix ladder for current
Common mistakes
- Moving the decimal in the wrong direction: mA to A makes the numerical value smaller by a factor of \(1000\).
- Mixing prefixes: \(\mu\text{A}\) (microampere) corresponds to \(10^{-6}\ \text{A}\), not \(10^{-3}\ \text{A}\).
- Unit inconsistency in formulas: current must match the unit system assumed by the equation (typically amperes in SI circuit equations).
Final answer
\[ 1\ \text{mA} = 10^{-3}\ \text{A} \qquad \Rightarrow \qquad I(\text{A}) = \frac{I(\text{mA})}{1000} \]