How can a multimeter detect damage to electronic products or home appliances?
This question is very complex, or in other words, it confuses me. For electronic product maintenance, home appliance repair, or computer brain maintenance, everyone has their own methods and cannot generalize. In general, a multimeter is just a tool with many functions (that is, it can measure many parameters). By proficiently measuring the parameters of various circuits or components, and combining them, it can determine the location of faults. For example, by measuring the voltage of a certain section of the circuit to determine whether it is working properly; Determine if a component is functioning properly by measuring its resistance; By measuring the capacitance of a component, determine whether it is open, short, or deteriorated.... wait. So a multimeter is just a tool. After DIY fans buy it back, they should still bubble up various forums or participate in extracurricular classes to strengthen their own literacy and improve their tool usage ability. Finally, to give an analogy, a multimeter is like a precious sword, and its power lies in its owner. With a certain percentage of martial arts (electronic knowledge), it can exert a certain percentage of its effectiveness.
How is the accuracy (uncertainty) of a digital multimeter calculated?
The accuracy of a multimeter, also known as uncertainty by some manufacturers, is generally stated as "within one year of leaving the factory, measured at an operating temperature of 18 ° C~28 ° C (64 ° F~82 ° F) and a relative humidity of less than 80%, ± (0.8% reading+2 characters) Many buyers or users are not very clear about this and often ask. I assume here that there is an instrument that, in a certain range, such as DC 200V, is written like this, and the measured value is displayed as 100.0 on the instrument. So, what should be the correct value at this time. I think for ordinary users, they can completely ignore accuracy calculations and simply assume that DC 100V is sufficient. According to the manufacturer's accuracy calculation, 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, use the displayed value to calculate. The calculated value should add the decimal point and then use the original reading to calculate the shipping cost. For example, the correct value is 100.0 ± 1.0, which should be between 99.0~101.0V DC.
