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Practical Quick Calculation Formulas for Electricians (1) I. Given the transformer capacity, determining the rated current at each voltage level: Formula: Capacity divided by the voltage value; the result is then multiplied by six and divided by ten. Note: Suitable for any voltage level. In daily work, some electricians are only involved in calculating the rated current of transformers for one or two voltage levels. By simplifying the above mnemonic, a mnemonic for calculating the rated current on each voltage level can be derived: multiply the capacity coefficients. II. Given the transformer capacity, quickly calculate the current values of the primary and secondary protection fuses (commonly known as insurance fuses). Mnemonic: For distribution transformers and high-voltage fuses, determine it by comparing capacity and voltage. For low-voltage fuses in distribution transformers, use capacity multiplied by 9 and then divided by 5. Note: The proper selection of fusing elements is of great significance for the safe operation of transformers. When fuses are used solely for protecting the high and low voltage sides of a transformer, the proper selection of the fuses becomes even more important. This is a problem that electricians often encounter and need to solve. III. Given the capacity of a three-phase motor, determine its rated current. Rule: Divide the capacity by the voltage in kilovolts; multiply the result by 0.76. Note: (1) This mnemonic is applicable to the calculation of the rated current of three-phase motors at any voltage level. Both formulas and mnemonics show that motors with the same capacity but different voltage levels have different rated currents; in other words, when the voltage in kilovolts varies, dividing by the same capacity yields different \"quotients.\" When these different quotients are multiplied by the same coefficient of 0.76, the resulting current values also differ. If the above formula is referred to as a general formula, then special formulas can be derived for calculating the rated current of motors operating at voltage levels of 220 V, 380 V, 660 V, and 3.6 kV. When using these special formulas to determine the rated current of a three-phase motor, the relationship between capacity in kilowatts and current in amperes is based on simple multiplication; there is no need to divide the capacity by the voltage in kilovols, nor to multiply the resulting figure by the coefficient of 0.76. Three-phase 220V motor, 3.5 kilowatts and 3.5 amperes. The commonly used motor is a 380V model, with one kilowatt and two amperes. Low-voltage 660 motor, 1.2 kilowatts and amps. 3,000-volt high-voltage motor, four kilowatts one ampere. High-voltage 6,000-volt motor, eight kilowatts one ampere. (2) When using the mnemonic, note that the unit for capacity is kW, the unit for voltage is kV, and the unit for current is A. This point must be kept in mind. (3) The coefficient of 0.76 in the mnemonic is a comprehensive value derived from calculations taking into account the motor’s power factor and efficiency, etc. The power factor is 0.85 and the efficiency is 0.9; these two values are more suitable for motors with a capacity of several dozen kilowatts, while they seem excessive for commonly used motors with a capacity of less than 10 kW. In this case, the rated current of the motor calculated using formula c differs from the value indicated on the motor’s nameplate; this difference has little impact on motors of 10 kW or less, as it affects the switch, contactor, wires, etc., based on the rated current. (4) Use mnemonic calculation techniques. To calculate the rated current of a common 380V motor using a mnemonic, first divide the motor’s supply voltage of 0.38 kV by 0.76; then multiply the resulting quotient by the motor’s capacity in kW. For larger 6kV motors, if the capacity in kW is exactly a multiple of 6kV, the capacity is divided by the voltage in kV, and the resulting figure is multiplied by a coefficient of 0.76. (5) Error. The coefficient of 0.76 in the formula is derived by assuming a motor power factor of 0.85 and an efficiency of 0.9; as a result, calculating the rated current for motors with different power factors and efficiencies leads to errors. The multiples of capacity (kW) and current (A) for the 5 specialized formulas derived from formula c are the quotient of the voltage level (kV) value divided by the 0.76 coefficient. Special mnemonics are simple for mental calculation, but it should be noted that their error margin increases. Generally, for units with a higher kilowatt rating, the calculated current is slightly higher than that indicated on the nameplate ； For those with a lower kilowatt rating, the calculated current is slightly lower than that indicated on the nameplate. Therefore, when calculating the current, if it reaches ten or several dozen amps, there is no need to consider values beyond the decimal point. It is possible to round off rather than use half-values, taking only whole numbers; this is simple and does not affect usability. For smaller currents, it is sufficient to consider one decimal place. Practical Electrical Engineer’s Quick Calculation Formulas (II) IV. Determining Capacity by Measuring Current 1. To estimate the rated capacity of a motor without a nameplate, measure its no-load current. Formula: For a motor without a nameplate, use the measured no-load current value, multiply it by ten and divide by eight to obtain an approximate value in kilowatts. Note: This mnemonic provides a method for estimating the capacity in kilowatts of a three-phase asynchronous motor without a nameplate, by measuring the motor’s no-load current. 2. Measure the current on the secondary side of the power transformer to determine its load capacity. Mnemonic: Given the secondary voltage of the distribution transformer, measure the current to find the power in kilowatts. 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. Note: (1) In their daily work, electricians often encounter situations where higher-level departments or managers ask about the operating conditions of power transformers and what the load level is The electrician himself often needs to know what the load on the transformer is. The load current is easy to determine; it can be seen directly from the ammeter installed on the power distribution unit, or measured using a corresponding clamp meter. However, the load power cannot be seen or measured directly. This requires the use of this mnemonic to carry out the calculations; otherwise, using conventional formulas would be both complicated and time-consuming. (2) “Voltage level: 400 volts, output power: 0.6 kilowatts.” "After measuring the load current on the secondary side of the power transformer (voltage level 400V), multiplying the amperage value by the coefficient of 0.6 yields the load power in kilowatts. 3. Measure the current in the wiring of an incandescent lamp to determine its load capacity; the lighting voltage is 220 volts, with 1 ampere corresponding to 220 watts. Note: Lighting in industrial and manufacturing enterprises typically uses 220V incandescent lamps. The lighting power supply circuit refers to the lines that run from the distribution panel to various lighting distribution boxes. The main lighting power supply lines are generally three-phase four-wire; a single-phase system can be used when the load is 4 kW or less. Lighting distribution circuits refer to the circuits that run from the lighting distribution box to lighting fixtures such as lights or sockets. Whether it is a power supply line or a distribution line, as long as the current value of a certain phase wire is measured using a clamp ammeter and then multiplied by a 220 factor, the resulting value represents the load capacity of that phase wire. Measuring current to determine capacity values can help electricians quickly address imbalances in the three-phase load capacity of lighting circuits. It also assists them in analyzing reasons why protective fuses in distribution boxes blow frequently, as well as the causes of overheating in power distribution wires. 4. To determine the rated capacity of a 380V single-phase welding transformer without a nameplate, use the following formula: The rated capacity is 380 multiplied by the no-load current, then divided by five. A single-phase AC welding transformer is essentially a step-down transformer for special purposes; compared to ordinary transformers, its basic working principle is roughly the same. 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; when the voltage drops to zero (that is, when there is a short circuit on the secondary side), 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**). This is the theoretical basis for mnemonics and formulas. V. Given the capacity of a 380V three-phase motor, determine the rated current and setting current of its overload protection thermal relay. Mnemonic: For protecting the motor against overload, use the thermal element of the thermal relay ； The flow capacity is 2.5 times, with the power set at twice the value. Note: (1) Motors that are prone to overload, may fail to start due to severe starting or self-starting conditions, or require limitation of the starting time, should be equipped with overload protection. Overload protection is also advisable for motors that run unmonitored for long periods or motors of 3 kW and above. Overload protection devices typically use thermal relays or time-delay overcurrent trip units in circuit breakers. The thermal relays produced in our country are currently suitable for overload protection of motors that start under light loads, operate for long periods of time, or experience intermittent long-term operation. (2) Thermal relay overload protection devices have very simple structural principles, but the selection of the adjustable heating element is delicate; if the appropriate rating is chosen too high, it must be adjusted to a lower value, which often leads to the motor stopping unexpectedly, affecting production and increasing maintenance work. If the rating is chosen too low, it can only be increased to the higher limit; as a result, the motor may not operate when overloaded, or even get damaged. (3) To correctly select the thermal relay for the overload protection of a 380V three-phase motor, it is also necessary to understand that thermal elements with different rated currents can be used in thermal relays of the same series model. The setting current for the heating element is determined as “twice the kilowatt rating”" ； The rated current of the heating element is determined by taking twice and a half the rated current capacity ； The model specifications of a thermal relay, that is, its rated current value, should be greater than or equal to the rated current value of the thermal element. VI. Given the capacity of a 380V three-phase motor, determine the rated current rating of its remote-controlled AC contactor. Rule of thumb: For a contactor used with a remote-controlled motor, the rating should be twice the motor’s capacity ； Step-by-step, start forward and reverse rotation, advancing one level at a time based on the current level. Note: The commonly used AC contactors at present include series such as CJ10, CJ12, and CJ20, which are suitable for controlling the starting of ordinary three-phase motors. VII. 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. Rule of thumb: For motors that are started directly, the capacity should 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: (1) The motor that can be started directly as mentioned in the mnemonic is a small 380V squirrel-cage three-phase motor; its starting current is very high, typically 4 to 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 infrequent direct starting of motors with a power rating of 10 kW or less. Both require a melt for short-circuit protection, and the motor power must not exceed 30% of the capacity of the supply transformer. In short, keep in mind that starting a motor directly with a load switch is conditional! (2) Load switches are all composed of simple isolating switch blades and fuses or fusing elements. To avoid the high current during motor startup, the capacity of the load switch, that is, its rated current (A) ； The rated current (in amps) of the fuse used for short-circuit protection is determined according to the rule that \"the switch should be selected based on six times the kilowatt rating, while the fuse should be chosen based on five times the kilowatt rating.\" Since both iron-clad switches and isolators with plastic covers and porcelain bases are manufactured to specific standards, the current value calculated using this rule still needs to be close to the specifications of the switch itself. Also considered a type of melt, it should be selected according to the product specifications. VIII. Given the capacity of a cage-type motor, calculate the operating time of a star-delta starter (QX3, QX4 series) as well as the setting current of its thermal elements. Mnemonic: For motor starting using the star-delta method, the starting time is easy to determine ； Take the square root of the capacity, multiply by two; add four units of seconds to the result. Motor start: star-delta, overload protection: thermal element ； The setting current for the phase current is capacity multiplied by eight divided by seven. Note: (1) The QX3 and QX4 series are automatic star-delta starters, consisting of three AC contactors, a three-phase thermal relay, and a time relay; they are equipped with a start button and a stop button. Before using the starter, the time relay and thermal relay must be properly adjusted, and both of these tasks are carried out at the site where the starter is installed. Most electricians only know the capacity of motors, but not their normal starting time or rated current. The operating time of a time relay is the start-up time of the motor (the time from startup until the speed reaches the rated value), and this value can be calculated using a mnemonic. (2) When adjusting the time relay, do not connect the motor for operation at this stage; instead, test whether the operating time of the time relay matches the start-up time of the motor being controlled. If there is a discrepancy, the operating time of the time relay should be fine-tuned again, and testing should be conducted once more. However, the interval between the two tests must be at least 90 seconds to ensure automatic reset of the bimetallic time relay. (3) Adjustment of the thermal relay: Since the thermal element in the thermoelectric device of the QX series starters is connected in series in the motor’s phase current circuit, and the motor is connected in a triangular configuration during operation, the phase current during motor operation is 1/√3 times the line current (i.e., the rated current). Therefore, the setting current value of the thermal element in a thermal relay should be calculated using the formula in the mnemonic: \"capacity multiplied by eight divided by seven\". Based on the calculated value, adjust the setting current knob of the thermal relay to the corresponding scale – around the midpoint scale. If the calculated value falls outside the rated current range of the thermal element in the thermal relay – that is, it is greater than or less than the upper or lower limit values indicated on the scale of the adjustment mechanism – then it is necessary to replace it with an appropriate thermal relay or select a suitable thermal element. IX. Given the capacity of a cage-type motor, determine the setting current for the circuit breaker’s trip device. Rule: The setting current for the circuit breaker’s trip device is equal to multiple times the motor’s capacity ； The instantaneous value is generally twenty; for smaller motors, it is twenty-four ； The time-delay trip is 3.5 times, while that of the thermal trip device is twice. Note: (1) Automatic circuit breakers are commonly used as circuit breakers that are not operated frequently in the circuits supplying power to squirrel-cage motors. If the operation is frequent, an additional contactor can be added for control. Circuit breakers use their electromagnetic trip units (for instantaneous protection) to provide short-circuit protection, and their thermal trip units (or time-delayed trip units) to provide overload protection. Calculating the setting current value for a circuit breaker’s trip device is a common issue faced by electricians; a mnemonic helps determine the multiplicative relationship between the setting current value and the capacity of the squirrel-cage motor in kilowatts. (2) “3.5 times for time-delay tripping, and twice for thermal trip” refers to the fact that for circuit breakers used as overload protection, the current setting for the time-delay tripping device can be chosen as 1.7 times the rated current of the motor being controlled; in other words, it should be set at 3.5 times the kilowatt value. The current setting value of the thermal trip device should be equal to or slightly greater than the rated current of the motor; in other words, it should be selected as twice the motor’s capacity in kilowatts. X. Given the capacity of an asynchronous motor, calculate its no-load current. Mnemonic: For the no-load current of a motor, use about 80% of its capacity as a reference ； New large models are 40% off, old small models have higher kilowatt ratings. Note: (1) When an asynchronous motor is operating under no-load conditions, the current flowing through its stator three-phase windings is called no-load current. The vast majority of the no-load current is used to generate a rotating magnetic field; this is known as the no-load excitation current, and it represents the reactive component of the no-load current. A very small portion of the no-load current is used to generate various power losses that occur when the motor is operating without a load (such as friction, ventilation, and core losses). This portion constitutes the active component of the no-load current; since its proportion is very small, it can be ignored. Therefore, no-load current can be considered as reactive current. From this perspective, the smaller it is, the better, as this improves the power factor of the motor, which is beneficial for power supply to the grid. If the no-load current is high, since the cross-sectional area of the wires in the stator windings is fixed and the current that can pass through them is also fixed, the active current that can flow through these wires must decrease. As a result, the load that the motor can drive decreases, and its output power drops. When too large a load is applied, the windings are prone to overheating. However, the no-load current cannot be too low either, otherwise it will affect other performance aspects of the motor. The no-load current of small electric motors is generally around 30% to 70% of the rated current, while that of large and medium-sized electric motors is about 20% to 40% of the rated current. The no-load current of a specific motor is generally not indicated on the motor’s nameplate or product manual. Electricians often need to know this value, as it is used to assess the quality of motor repairs and determine whether the motor is suitable for use. (2) The mnemonic is a formula for quickly calculating the specific value of the motor’s no-load current on-site; it is derived from numerous test data. It conforms to the principle that \"the no-load current of a motor is generally 1/3 of its rated current.\" At the same time, it is in line with practical experience: the principle that \"the no-load current of a motor should not exceed its capacity in kilowatts in order for it to be usable\" (applying to old, low-capacity motors after maintenance). The mnemonic \"Calculate it at about 80% of the capacity\" means that the no-load current value of a typical motor is roughly 0.8 times the motor’s rated capacity in kilowatts. The no-load current of medium-sized, 4- or 6-pole motors is 0.8 times the motor’s capacity in kilowatts ； New series: high-capacity, 2-stage motors with a relatively low number of poles; the no-load current is calculated at 60% of the value applicable to motors with a higher number of poles" ； For old, vintage series of motors with smaller capacity and a high number of poles – 8 poles or more – the no-load current is calculated based on the principle of \"fewer poles equal higher kilowatt rating\"; in other words, the no-load current value is approximately equal to the motor’s capacity in kilowatts, though it is generally less than that value. The mnemonic formula can be used to calculate the no-load current of a motor. The value obtained using this formula differs somewhat from the value indicated in the motor’s manual as well as from the actual measured value, but it is more than sufficient for the daily work needs of electricians. 11. Given the capacity of an electrical power transformer, determine the setting current value for the instantaneous trip device of the circuit breaker on its secondary side (0.4 kV). Mnemonic: For power supply on the secondary side of a transformer, it is advisable to use a circuit breaker ； Instantaneous trip setting value: three times the capacity in kVA. Practical Electrical Worker’s Quick Calculation Formulas (III) 12. Making Smart Use of Low-Voltage Voltage Testers Low-voltage voltage testers are a common auxiliary safety tool used by electricians. Used to check whether conductors below 500V or the casings of various electrical devices are charged. An ordinary low-voltage voltage tester can be carried around; as long as one understands its working principle and combines it with well-known electrical principles, there are many techniques that can be applied flexibly. 1. Mnemonic for distinguishing AC from DC: Use a test pen to identify them; AC results in bright light while DC produces dim light. In the case of neon tubes, AC causes the entire tube to glow, whereas DC makes only one end of the tube bright. Note: First, inform the reader that before using a low-voltage voltage tester, it is necessary to test it on a known live conductor first ； It must not be used until it is confirmed that the voltage tester is functioning properly. To distinguish between alternating 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. 2. Mnemonic for determining the positive and negative poles of direct current: Use a test pen to identify the poles; be careful when observing the neon tube. If the front end is bright, it is the negative pole, while if the back end is bright, it is 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, by touching one pole of the power source with one hand and holding a test pen in the other hand, with the metal tip of the pen in contact with the other pole of the power source under test, the tip of the neon tube will light up; in this case, the pole being touched is the negative pole ； If the rear terminal of the neon tube is glowing, 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 to the positive pole. 3. Determine whether the DC power supply is grounded; there is a mnemonic to distinguish between grounding of the positive and negative poles: For the DC system in a substation, the test pen does not light up when it makes contact with such a ground ； If the light is close to the tip of the pen, there is a ground fault at the positive pole ； If the light is near the tip of the finger, the ground fault is on the negative pole. Note: The DC coefficients of power plants and substations are isolated from the ground. When a person stands on the ground and uses a voltage tester to touch the positive or negative pole, the neon tube should not light up; if it does light up, it indicates that there is a ground fault in the DC system ； If the glow is at the end near the pen tip, the positive pole is grounded ； If the glow is at the end near the finger, the negative pole is grounded. 4. Mnemonic for determining in-phase and out-of-phase states: To determine whether two wires are in the same phase or not, hold a pen in each hand; keep your feet insulated from the ground. Touch each wire with one of the pens, then look at one of the pens – 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 strokes appear the same whether lit or not, so it is sufficient to check just one of them. 5. Mnemonic for identifying phase-to-ground faults in 380/220V three-phase three-wire supply systems: In a star connection, when a test pen touches two of the phase wires, they will light up; the remaining wire will show a weaker glow, indicating that that phase wire is grounded ； If there is almost no light, it is a fault with the metal grounding. Note: The secondary side of a power transformer is generally connected in a Y configuration. In a three-phase three-wire system with an ungrounded neutral point, when using a voltage tester to touch the three phase wires, two of them will emit a slightly brighter light than usual, while the third wire will show a weaker brightness. This indicates that there is a grounding issue with that wire, although it is not severe yet ； If two wires are very bright while the remaining one is almost invisible in brightness, it indicates that there is a metal-to-ground fault in that live wire.