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This mnemonic summarizes common electrical engineering theories, data, construction procedures, and the usage methods of instruments and equipment. It is called an electrical worker’s mnemonic because it is inspired by the ease with which poetry can be spread and remembered. The words in the mnemonics are generally neat, concise, fluent, and rhyming, which makes the otherwise dull electrical engineering theories and data easier to remember. These 100 mnemonics are merely a starting point; what we hope more is that our colleagues will share their experiences and insights from practice, as well as collect similar mnemonics used in actual work, so as to make them more comprehensive and useful. Electrical Engineer’s Mnemonic (1): A simple way to estimate the current-carrying capacity of wires. For wires below 10 mm2, the capacity is 5 A per square millimeter; for 100 mm2 and above, it’s 2 A per square millimeter. There are three ranges at 25, 35, and 55 mm2. For 70 and 95 mm2, the capacity is 2.5 times higher. The actual capacity should be reduced by 10-15% due to temperature effects, and the capacity of copper wires increases when using higher-grade copper materials. Explanation: For aluminum wires with a cross-sectional area of 10 mm2 or less, the current-carrying capacity is calculated as 5 A per square millimeter ； The current-carrying capacity of aluminum wires of 100 mm2 and above is calculated at 2 A per square millimeter ； The current-carrying capacity of 25mm2 aluminum wires is calculated at 4A per square millimeter ； The current-carrying capacity of 35mm2 aluminum wires is calculated at 3A per square millimeter ； The current-carrying capacity of 70mm2 and 95mm2 aluminum wires is calculated at 2.5A per square millimeter ； \"Copper material upgrade calculation\": For example, to calculate the current-carrying capacity of a 120mm2 copper wire, an 150mm2 aluminum wire can be used, and the current-carrying capacity of the aluminum wire is then determined ； Affected by temperature, a factor of 0.8 or 0.9 (depending on the location) must also be multiplied in. Electrical technician’s mnemonic (2): Given the transformer capacity, determine the rated current on the voltage level side. Note: This applies to any voltage level. Mnemonic: Capacity divided by voltage value; the result multiplied by six and then divided by ten. Example: Apparent current I = Apparent power S / 1.732 × 10^KV = 1000 KVA / 1.732 × 10^KV = 57.736 A. An estimated value is I = 1000 KVA / 10^KV × 6/10 = 60 A. Electrician’s mnemonic (3): Given the transformer capacity, quickly determine the current values of the primary and secondary protection fuses (commonly known as insurance fuses). Mnemonic: For high-voltage fuses in transformers, calculate using the ratio of capacity to voltage. For low-voltage fuses in distribution transformers: the capacity is multiplied by 9 and then divided by 5 – an electrical formula mnemonic (4). To find the rated current of a three-phase motor, use this mnemonic: divide the capacity by the voltage in kilovolts, and then multiply the result by 0.76. It is known that the three-phase 220V motor has a power of 3.5 kilowatts and a current of 3.5 amperes. 1KW÷0.22KV*0.76≈1A A motor with a high voltage of 3,000 volts consumes 4 kilowatts per ampere. 4KW÷3KV*0.76≈1A Note: This formula can be used to calculate the rated current of three-phase motors at any voltage level. When using the mnemonic, the capacity unit is kW, the voltage unit is kV, and the current unit is A. Electrician’s Mnemonic (5): To determine the current on the secondary side of a power transformer and thus calculate the load capacity it can handle, if the secondary voltage of the transformer is known and the current is measured, then the power in kilowatts can be calculated. Voltage level: 400 volts, 1 ampere, 0.6 kilowatts. Voltage level: 3,000 volts; power: 1 ampere at 4.5 kilowatts. Voltage level: 6,000 volts; power: exactly 9 kilowatts per ampere. Voltage level: 10 kV; current: 1 ampere; power: 15 kilowatts. Voltage level: 35 kV, 1 ampere at 55 kilowatts. Electrician’s Mnemonics (6): Given the capacity of a small 380V three-phase squirrel-cage motor, determine the minimum capacity of the power supply equipment, as well as the current values for the load switch and protective fuses required to start the motor directly, provided that its capacity does not exceed 10 kilowatts ； 6-kilowatt selector switch, 5-kilowatt fuse. For power supply equipment in kVA, the kilowatt rating needs to be three times higher. Note: The electric motor referred to in the mnemonic is a small 380V squirrel-cage three-phase motor. Its starting current is very high, typically 4-7 times the rated current. The capacity of motors that can be started directly using a load switch should not exceed 10 kW; generally, 4.5 kW or less is advisable. Open-type load switches (isolating switches with rubber covers and porcelain bases) are typically used for small-capacity motors of 5.5 kW or less, for infrequent direct starting ； Closed-load switches (iron-clad switches) are generally used for the infrequent direct starting of motors with a power rating of 10 kW or less. Load switches consist of simple isolating switch blades along with fuses or fusing elements; it is advisable to choose a switch with a rating that is 6 times the required power ； To avoid the high current that occurs when the motor starts, it is advisable to use a fuse with a rating five times that of the motor’s rated power, that is, the rated current (A) ； Rated current of the fuse for short-circuit protection (A). Finally, an appropriate power supply must be selected, with its output power being no less than 3 times the rated power. Electrician’s mnemonic (7): To determine the rated capacity of a 380V single-phase welding transformer without a nameplate, use this mnemonic: For a welding transformer with a voltage of 380V, multiply the no-load current by five to get the rated capacity. A single-phase AC welding transformer is essentially a step-down transformer for special purposes; its basic working principle is roughly the same as that of ordinary transformers. To meet the requirements of the welding process, the welding transformer operates in a short-circuit condition, and it is necessary to have a certain arc-starting voltage during welding. As the welding current increases, the output voltage drops sharply. According to P=UI (with power constant, voltage is inversely proportional to current). When the voltage drops to zero (that is, when the secondary side is short-circuited), the current on the secondary side does not become excessive either; in other words, the welding transformer has a steep voltage-drop characteristic, and this characteristic is achieved thanks to the voltage drop generated by the reactance coil. When unloaded, since no welding current flows through it, the reactance coil does not generate a voltage drop; at this time the unloaded voltage is equal to the secondary voltage. In other words, the behavior of a welding transformer when unloaded is the same as that of an ordinary transformer when unloaded. The no-load current of a transformer is generally around 6% to 8% of the rated current (**it is specified that the no-load current should not exceed 10% of the rated current**). Electrician’s Mnemonic (8): Determining AC from DC current. Use a test pen to identify AC and DC: AC results in bright light, while DC produces dim light. In the case of AC, the neon tube lights up entirely; for DC, only one end of the neon tube lights up. Note: When distinguishing between alternating current and direct current, it is best to make a comparison between the two types of electricity, which makes it clear. When measuring alternating current, both ends of the neon tube glow; when measuring direct current, only one end of the neon tube glows. Electrician’s Mnemonic (9): Using a test pen skillfully to determine phase alignment at low voltage. To check whether two wires are in the same phase or not, hold one test pen in each hand; ensure that both feet are insulated from the ground. Touch each wire with one of the pens, then observe the pen – if it does not light up, the wires are in the same phase; if it does light up, they are in different phases. Note: During this test, make sure both feet are insulated from the ground. Since most areas in our country are powered at 380/220V, and transformers typically have their neutral points directly grounded, it is essential to ensure insulation between the human body and the ground during testing, in order to prevent the formation of a circuit and avoid incorrect readings ； During testing, the two lights appear the same whether on or off, so it is sufficient to check just one of them. Electrician’s Mnemonics (10): Using a test pen to determine the positive and negative poles of direct current. To use a test pen to identify the poles, one must pay close attention to the neon tube: if the front end is bright, it indicates the negative pole, while if the back end is bright, it indicates the positive pole. Note: The front end of the neon tube refers to the tip of the voltage tester, while the back end refers to the end held in the hand. If the front end is bright, it is the negative pole; otherwise, it is the positive pole. During testing, note that the power supply voltage should be 110V or higher ； If a person is insulated from the ground, and one hand touches one pole of the power source while the other hand holds a voltage tester whose metal tip comes into contact with the other pole of the power source, the bulb at the front end of the tester will light up; in this case, the pole being touched is the negative pole ； If the rear pole of the neon tube is glowing, then the power source being tested is the positive pole, based on the principle of one-way flow of direct current and the flow of electrons from the negative pole to the positive pole. (To be continued) More articles on the applications of electrical measuring instruments are available at: http://www.fluke.com/fluke/cnzh/Support/appnotes/default.htm