How does a multimeter detect damage to electronic products or household appliances?
This question is very complicated, or I was overwhelmed by the question. When it comes to electronic product maintenance, home appliance repair or computer maintenance, everyone has their own methods and cannot be generalized. In general, a multimeter is just a tool with many functions (that is, it can measure many parameters). By skillfully measuring the parameters of each circuit or component, and combining them, the fault location can be determined. For example, by measuring the voltage of a certain circuit to determine whether it is working properly; by measuring the resistance of a component to determine whether it is normal; by measuring the capacitance of a component to determine whether it is open circuit, short circuit or deteriorated. . Therefore, a multimeter is just a tool. After you buy it, DIY fans should also visit various forums or participate in amateur classes to strengthen your own literacy and improve your ability to use the tool. Finally, to use an analogy, a multimeter is like a sword. Its power lies in its owner. How much martial arts (electronics knowledge) there is, how much effectiveness it can exert.
How is the accuracy (uncertainty) of a digital multimeter calculated?
The accuracy of a multimeter is also stated by some manufacturers as uncertainty, which generally states "within one year after leaving the factory, when measured at an operating temperature of 18°C to 28°C (64°F to 82°F) and a relative humidity of less than 80%, ± (0.8% reading + 2 words)." Many buyers or users are not very clear about this and often ask. I am assuming here that there is an instrument that is in a certain range, such as the DC 200V range, and it is written like this. The measured value displays 100.0 on the instrument, so what should the correct value be at this time. I think that for ordinary users, they can completely ignore the accuracy calculation and just think of it as DC 100V. Calculated according to the manufacturer's accuracy, when measuring 100V (displaying 100.0), the error is ±(0.8%*1000+2)=±10, which is an error of 1.0V. When substituting the reading, do not consider the decimal point to display Substitute the value into the calculation. Add a decimal point to the calculated value and then use the original reading to calculate the shipping cost. Like this example, the correct value is 100.0±1.0, which should be between 99.0" and 101.0V DC.
