What is the difference between high precision resistors and current sense resistors
Resistors are resistors. They only see the current thru them and the voltage across them.
However, specific models can be targeted to specific applications. What you show as a current sense resistor looks like it's designed to dissipate significant power. However, it would work for lots of purposes within its power and voltage limits.
Current sense resistors tend to:
- Be run with little voltage across them.
- Be fairly accurate, since being used for measurement.
- Drift little due to temperature.
- Have low values.
- Sometimes have 4 leads so that you can use a Kelvin connection. Two leads carry the current, then the voltage is measured across the two other leads. This keeps the voltage drop due to current in the leads off the measurement.
There is no absolute difference between the two.
Current sense resistors are generally low-value, high-power resistors. They are intended to develop a small voltage for a given current, so they are low in value. When used in power supplies they may well carry significant current and dissipate significant power, so they tend to be high-power units. For many applications they are not used to measure current with high precision, but this is not guaranteed.
Precision resistors are simply resistors which have stable, well-defined resistance. The existence of non-zero temperature coefficients means that, for a similar size a very high precision resistor will be limited to lower power than a non-precision resistor, although this does not apply for precisions of 1 % or less.
Your picture of a current sense resistor shows a unit which (if you look at it closely) has 1 % precision. The other picture shows resistors with 5% precision (the gold band at left), so they are lower-precision than the current-sense resistor which you show.
In your case, you need to measure 30 mA. What voltage are you willing to drop across the resistor at this current? Let's assume that you want 0.1 volts. Then the resistor should have a value of $$R = \frac{V}{i} = \frac{0.1}{0.03} = 3.33 \text{ ohns} $$ and at maximum power it will dissipate $$P = i^2 R = (.03)^2 \times 3.33 = 3\text{ mW} $$
As a result, if 0.1 volts is acceptable, you can use a standard 1/10 watt, 1% resistor.