- Basic electrical theory tells us that for a single-phase system, kW = (V × I × PF) ÷ 1,000. For the sake of simplicity, let's assume the power factor (PF) is unity. Therefore, the above equation become
- Single vs. Three Phase AC - Amperage - Convert between single phase (120, 240 and 480 Voltage) and three phase (240 and 480 Voltage) Three Phase Electrical Motors - Full load amps, wire and conduit sizes for three phase electrical motors; Three-Phase Power Equations - Electrical 3-phase equation
- In electrical engineering, single-phase electric power is the distribution of alternating current electric power using a system in which all the voltages of the supply vary in unison. Single-phase distribution is used when loads are mostly lighting and heating, with few large electric motors. A single-phase supply connected to an alternating current electric motor does not produce a rotating.

Reactive Power Of Single & 3-Phase Current: Where. θ = is the phase angle i.e. phase difference between voltage and current. The following table shows the different electrical power formulas in both AC and DC circuits Single & Three Phase Line kVA calculator is an online tool used in electrical engineering to measure the unknown quantity by two known quantities applied to the below formulas for single phase and three phase connection. To calculate kVA, you need to enter the known values of voltage and the current into the respective fields Formula ; Simple electrical calculator to calculate one phase power in a circuit based on voltage and current. Single phase power is the supply of single alternating current runs through a single circuit. It is the most common household power circuit used in our day to day life. Single phase AC power systems have high peak voltage that lies. NOTE: Direct Current formulae do not use (PF, 2, or 1.73); Single phase formulas do not use (2 or 1.73); Two phase-four wire formulas do not use (1.73); Three phase formulas do not use (2) * For three-wire, two phase circuits, the current in the common conductor is 1.41 times the current in either of the other two conductors Example 1: Find the apparent power in kVA drawn by a 200 V single phase generator having 2 Amps current passing through it. Solution: kVA = 200 * 2 / 1000 = 0.4 kVA Example 2: Find the current drawn by 500 kVA transformer @ rated voltage 5 kV

Single phase to 3 phase power calculation input requirement = The square root of 3 (1.732) x 10 amps = 1.732 x 10 amps = 17.32 Amps Single phase power input in this instance is 17.32 Amp Single Phase Transformer Full-Load Current (Amps)= kVA × 1000 / V Three Phase Transformer Full-Load Current (Amps) = kVA × 1000 / (1.732 × V The power taken by a circuit (single or three phase) is measured in watts W (or kW). The product of the voltage and current is the apparent power and measured in VA (or kVA) . The relationship between kVA and kW is the power factor (pf): which can also be expressed as: Single phase system - this is the easiest to deal with. Given the kW and. This video explains the formula to calculate current in single phase circuits when HP, kW or watts are provided

Put the details obtained in Step 1 in the following formula . For 3 phase transformers . Let's put the values we assumed in Step 1 in above formula. For 1 phase transformers . So, these are the two simple steps you can use to calculate the primary current of any single phase or three phase transformer Electrical Formulas; Fuel Consumption; Amperage Chart; Search for: Formula Chart. Standard Electrical Formulas Used for Power Consumption Calculations. TO DETERMINE: SINGLE-PHASE: THREE-PHASE: DIRECT CURRENT: KVA: I x E 1000: I x E x 1.73 1000-----Kilowatts: I x E x PF 1000: I x E x 1.73 x PF 1000: I x E 1000: Horsepower: I x E x %EFF x PF 746. You can calculate the line current of a transformer by using the appropriate formula for single-phase or 3-phase systems: Single-phase: I = VA ÷ E. 3-phase: I = VA ÷ (E × 1.732) Overcurrent protection. To protect the windings of a transformer against overcurrent, use the percentages listed in Table 450.3(B) and its applicable notes Short-Circuit Current Calculations Single-Phase Short Circuits Short circuit calculations on a single-phase center tapped transformer system require a slightly different procedure than 3Ø faults on 3Ø systems. 1. It is necessary that the proper impedance be used to represent the primary system

AC single phase watts to amps calculation formula The phase current I in amps (A) is equal to the real power P in watts (W), divided by the power factor PF times the RMS voltage V in volts (V): I(A) = P(W) / (PF × V(V)) So amps are equal to watts divided by power factor times volts Download and print Single Phase and Three Phase AC Power Chart. Example - Power Supply to Electric Heater. 10 kW of power is required by an electric heater. The power supply available is 230V single or three phase. From the chart above we can estimate the current in the two alternatives to be approximately. 43 A with single phase 230V; 25 A. three phase power is 36 kW, single phase power = 36/3 = 12 kW now simply follow the above single phase method kVA = kW / power factor = 12 / 0.86 = 13.9 kVA (13900 VA) Current = VA / voltage = 13900 / 230 = 60

- There is no way to calculate the Starting Current or Locked Rotor Current (LRA) without more information! Single-phase or three-phase? NEMA Motor Design B, C or D? What does academic education's sake mean? A voltage of 15V with a power of 132kW is meaningless for an induction motor. You just can't make up numbers
- One phase and neutral are used together to give a single-phase power supply. The 1-phase power supply uses AC power (alternating current) to be distributed to the domestic householders. In a 1 phase supply, the voltage varies sinusoidally with respect to the time period as shown in the figure
- AC single phase kilowatts to amps calculation The phase current I in amps (A) is equal to 1000 times the power P in kilowatts (kW), divided by the power factor PF times the RMS voltage V in volts (V): I(A) = 1000 × P(kW) / (PF × V(V)) AC three phase kilowatts to amps calculatio
- In a single-phase (or 3-phase for that matter) AC electrical power distribution system, both the load current (as electrical utilization loads are connected) and the applied voltage that causes it to flow are transmitted along the length of the circuit conductors varying in amplitude with time and periodically reverse between positive and negative polarities

Basic Voltage, Current, Power and Resistance Formulas in AC and DC Circuits. Following are the electrical engineering formulas and equations for the basic quantities i.e. current, voltage, power, resistance and impedance in both DC and AC circuits (single phase and three phase) In a single-phase system, the neutral carries only the imbalance of the two hot conductors, an easy calculation. However, in a three-phase wye system, even if only two of the three phases and the neutral run a single-phase load, you must use the neutral formula **Formula** to calculate power, **current** and voltage in three-**phase** wiring (unbalanced load, different loads on each of three **phases**) : Pt=P1+P2+P3 P1=V*I1*cosφ1 I1=P1/(V*cosφ1) Same thing for each **phase** V=P1/(I*cosφ1) Pt = total power of the circuit in watt (W) P1, P2, P3 = power of **phase** 1, **phase** 2 and **phase** 3 in watt (W

This motor full-load amperage (FLA) calculator allows you to calculate the full-load current of the AC electric motor. Instructions: Select the number of phases from the drop-down list; The motor FLA calculator uses the following formulas: Single Phase AC Motor FLA (Amperes) = (P [kW] × 1000) / (V × cos ϕ Let's take the condition first where the single phase power circuit is fully resistive in nature, that means the phase angle between voltage and current i.e. φ = 0 and hence, From the above equation it is clear that, whatever may be the value of ωt the value of cos2ωt cannot be greater than 1; hence the value of p cannot be negative

using a one step formula F p1 =0.7 F p2 =0.95 P T =10,000 W Ans Power Factor and Load Current 20 lesson 3_et332b.pptx Example 3-3: A 480 V, 60 Hz, single phase load draws 50.25 kVA at a power factor of 0.87 lagging. Find: a) the current and the active power in kW that the load absorbs b) the angle between the source voltage and the load current ** Formula for calculating single-phase and three-phase short circuits of the transformers (kA): VA = Volt ampere or active power**. Volts = Volts of the transformer. % Impedance = Impedance of the transformer . How to calculate the short-circuit level of a transformer in 1 single step: Step 1

The current is maximum in the beginning of the cycle and becomes zero at the maximum value of the applied voltage, so there is a phase difference of 90° between the applied voltage and current. During the first quarter-cycle the current flows in the normal direction through the circuit; hence the current is positive This can be calculated from primary or secondary winding voltage and current information. KVA is calculated by multiplying the line to line maximum voltage & maximum phase current; then multiplying this value with 1.732 in case of three phase ; lastly dividing the final value by 1000 for KVA and 100,000 for MV

- Single phase AC has the advantage that it only requires 2 wires. Its disadvantage is seen in the graph at the top of this page: twice every cycle V goes to zero. If you connect a phototransistor circuit to an oscilloscope, you will see that fluorescent lights turn off 100 times per second (or 120, if you are on 60 Hz supply)
- Thus for faulted phase, in case of Single Line to Ground Fault, we can write. I a0 = I a1 = I a2 = I a /3. Calculation of Fault Current: First thing which must be understood at this point, that fault current is completing its path through the grounded neutral. If there were no any grounded neutral, no fault current would have been flow
- The phase voltages are all equal in magnitude but only differ in their phase angle. The three windings of the coils are connected together at points, a 1, b 1 and c 1 to produce a common neutral connection for the three individual phases. Then if the red phase is taken as the reference phase each individual phase voltage can be defined with respect to the common neutral as
- KVA & Power Formulas: Power Formula-DC: KVA | Power Formula-Single Phase: KVA | Power Formula-3 Phase: Power(watts) = Voltage x Current W = E x I Power(watts) = Voltage² / Resistance W = E² / R Power(watts) = Current² x Resistance W = I² x
- As a rules of thumb amps horsepower rating can be estimated to. 115 volts motor - single-phase : 14 amps/hp; 230 volts motor - single-phase : 7 amps/hp; 230 volts motor - 3-phase : 2.5 amps/hp; 460 volts motor - 3-phase : 1.25 amps/hp; Always check nameplate information before designing protective devices, wiring and switch gear
- Single-phase full-wave diode rectifier. Single-phase diode rectifier, converting ac signal into a dc voltage, exist in two types - half-wave and full-wave one. Half-wave diode rectifier was mentioned before. Full-wave diode rectifier can be two types as well - with a centre-tapped transformer and bridge rectifier
- Phase to Neutral is 220V. I want to install a single phase elevator that has the following specs: Voltage = 220V. Power = 3700Watt. If I calculate the current requirement with the following formula: I = P/V I get 16.81A (3700W/220V). I called the sales to reconfirm and they told me that the required current to operate the elevator is 25A

AC Power is an alternating current electrical energy consumed by the load connected in an electrical circuit, generally measured in horsepower, watts, kilo-watts or kilo-watts-hour. Whereas, single phase electric power is the electrical energy consumed by the load from phase & neutral.Three phase electric power is also the measure of electrical energy consumed by the load from three diffrent. To calculate the Maximum Fault Current that can be achieved in a circuit fed by a transformer we refer to the below formula. This formula calculates the loop impedance at the end ofa circuit fed from the secondary winding of a transformer. When calculating Fault Loop Impedance Select the Correct Transformer Type Above. Three Phase Transformers. The Locked Rotor Torque or Starting Torque is the torque the electrical motor develop when its starts at rest or zero speed. The magnitude of the Locked rotor **current** is determined by motor horsepower and design characteristics. Normally it will be 6 to 7 times of the Motor full Load **current** First we must convert from hp to kW using this formula : 1 Hp = 0.746 kW. After that use the formula kW to ampere : For 3 phase power supply ; kW = I x V x 1.732 x pf. For 1 phase power supply ; kW = I x V x pf . Example :-1 unit induction motor power rated 25 hp,200 VAC,3 phase,90 % power factor,calculated the full load ampere. Calculation : locked rotor current is calculated during design.It depends upon so many factors like rotor bar configuration,Induct ance of windings on stator ,Flux density and so on.It is approximately 6 times the Full load current of the motor on DOL starting.The satrting current values are subject to tolerance of 20%.You can peruse any catalogue of motor manufcaturer and see the starting current given in.

A single-phase vacuum has an AC voltage of 127 volts (LN), 4.3 Amps and a power factor of 0.92, how many watts does the vacuum have? Rta: // The first thing to do is identify the formula to be used, because the equipment is single-phase and AC (alternating current), you should use the formula: V (LN) xIxF.P, replacing the variables would be. * A single phase system consists of just two conductors (wires): one is called the phase (sometimes line*, live or hot), through which the current flows and the other is called neutral, which acts as a return path to complete the circuit 265V / 460 V, 3 Phase, WYE, Y 2. Current (I) Current is the flow of electrons through a load, the units are ampers or amps. Current is measured inductively with a clamp-on ammeter. Single Phase Current ( I ) = (Kilovolt Amps) = KVA Kilovolts KV Example: Load = Unit heater 5 KW, 230V 1 phase

- To calculate the total power output of a motor, we can use the formula: HP = I x V x Eff / 746. I = Ampere. V = voltage. Eff = efficiency (available on the motor name plate) Example : 1 unit single phase motor,240 volt,20 ampere and 85% efficiency. HP = I x V x Eff / 746. HP = 20 x 240 x 85% / 746. HP = 5.
- The formula to find amps for a three-phase AC circuit is a little different from the formula for a single phase circuit: For example , find the current of a 25kW three-phase generator with an efficiency of 80% at 240 volts
- Single phase power system schematic diagram shows little about the wiring of a practical power circuit. Depicted above, is a very simple AC circuit.If the load resistor's power dissipation were substantial, we might call this a power circuit or power system instead of regarding it as just a regular circuit

Phase to Phase Fault In a phase to phase fault (L1 to L2 for example), two phases are connected together. The fault current is again, the nominal applied voltage divided by the summed impedance. Earth Faults. Earth Fault In an earth fault, one phase is directly connected to earth (L1 to earth for example) Single phase kW to Amps. The single phase and all other alternating current circuits introduce an additional concept of power factor in the denominator. Single phase formula is a ration of thousand times kilowatts to the power factor times operating voltages. Mathematically. I = [kW * 1000] / [E * PF] Example: A 120-volt single phase AC circuit. * Circuit diagram of single line-to-ground fault*. Suppose the phase a is connected to ground at the fault point F as shown in a figure below. I a, I b and I c are the current and V a, V b and V c are the voltage across the three phase line a, b and c respectively. The fault impedance of the line is Z f.. Since only phase a is connected to ground at the fault, phase b and c are open circuited and.

supplied with 10 AWG copper. The motor has a full load current draw of 24 amps. What is the voltage drop when the motor is in operation? (Please note that for motors or continuous loads, the total current of the load in question is at 100%, not at 125%). Answer: Applying the single-phase formula for voltage drop, where: K = 12.9 ohms-cmil/f A single-phase transformer is an electrical device that accepts single-phase AC power and outputs single-phase AC. This is used in the distribution of power in non-urban areas as the overall demand and costs involved are lower than the 3-phase distribution transformer Test single-phase circuits by clamping the phase and neutral conductor. The measured value will be any current flowing to ground. Test three-phase circuits by clamping around all three-phase conductors. If a neutral is present, it should be clamped along with the phase conductors. The measured value will be any current flowing to ground Where, f is the supply frequency, R is the load resistance and RF stands for the ripple factor.The waveforms obtained for the diode rectifier are shown in Fig.4. (a) Fig.4. Waveforms of (a) Supply voltage and supply current (b)Output Voltage . It is observed that the nature of current of the single phase

- single phase circuits which happen to have peaks and valleys separated by a period of time. Following is a brief review of the principles involved in single phase power. 2A. Single Phase Voltage and Single Phase Current In general, the instantaneous voltage in a single phase AC electrical circui
- Diode Current Equation Working of Single Phase Full Bridge Inverter. The working of single-phase full-bridge using RLC load inverter can be explained using the following scenarios. Overdamping and Underdamping. From graph at 0 to T/2 if we apply DC excitation to RLC load. The output load current obtained is in the sinusoidal waveform
- Moreover, the single-phase voltage drop formula is given as. VD = \(\frac{2 LRI}{1000}\) And three-phase voltage drop formula is given as. VD = \(\frac{2 LRI}{ 1000 \times 0.866}\) Here, L = refers to the length of the circuit R = refers to the resistance in Omega (\(\Omega\)) I = refers to the load current in amperes. Solved Example on Voltage.
- ate any negative voltage occurrence at the load ter
- e the amps in single phase 1 HP, 120-volt motor having an efficiency rating of 88% operating at a power factor 0.9. Solution: Current = [1 * 746] / [120 * 0.88.

Full Load Current Calculator Parameters. Voltage (V): Specify the phase-to-phase V LL voltage for a 3-phase AC supply in volts. Specify the the phase-to-neutral V LN voltage for a 1-phase AC or DC supply. Select the phase arrangement: 1 Phase AC, 3 phase AC, or DC. Load (S): Specify the the load in kW, kVA, A, or hp Calculation: Find the starting current of 10 hp, three phase 220 V A class motor. Solution: The maximum kVA/hp for A class motor is 3.15 (code factor). S start = 10 hp * 3.15 = 31.5 kVA. I L = 31.5 kVA / (√3 * 220) = 82.66 A. You might want to learn: How to calculate Current from H In this video we explain transformer full load calculation for single phase and three phase both easilyBasically if have any plant than we should know how mu..

The 3 phase load current amps formula is explain with a three phase load voltage induction motor nameplate data. From this post learn complete about how to calculate the load current of 3 phase motor. Electricalonline4u A platform to learn electrical wiring, single phase, 3 phase wiring, controlling, HVAC, electrical installation, electrical. After the power has been converted to watts, use the Watt's Law formula to find current in amps. Amps = Watts ÷ Volts. Convert the amps to watts with our electrical conversion calculator, plus learn the formulas for DC, single-phase AC, and three-phase AC circuits Welcome to schneider electric's fault current calculator. Please select system type from these two This means our calculation is wrong somewhere, and that is; we took into consideration the formula of single phase transformer and not of three phase transformer. So now, if we calculate by using the formula of eq. (2), we get. kVA rating = √3 x 11 x 5.25 = 100 kVA for HV side. kVA rating = √3 x 415 x139= 99.99 kVA for LV side Electrical Formulas To Find Single Phase Two Phase -Four Wire Three Phase Amps when HP is known HP x 746 V x %EFF x PF HP x 746 V x %EFF x PF x 2 HP x 746 V x %EFF x PF x 1.73 Amps when KW is known KW x 1000 V x PF KW x 1000 V x PF x 2 KW x 1000 V x PF x 1.73 Amps when kVA is kVA x 1000 V kVA x 1000.

- The calculator utilizes various formulas to perform the calculations; Computation of Direct Current Kilowatts to amps I (A) = 1000 x P (kW) / V (V), which means that the current in amps is calculated by multiplying 1000 by the power in Kilowatts dividing the result by the voltage in volts. Calculation of AC Single phase Kilowatts to amp
- g a three phase transformer or a single unit three phase autotransformer can be built. The three phase autotransformers (see Fig-D) are connected in star-star(Wye-Wye). If the autotransformers are connected as Delta-Delta, then phase difference between primary and secondary exist which.
- g a 3-phase power calculation means using a simple formula relating the power to the voltage, current and a power factor describing the efficiency of the circuit. The key difference between working out power for 3-phase and single-phase systems is a factor of the square root of three
- e the correct way to calculate the current in a 480V, single phase circuit, which services ten roadway lighting fixtures. The circuit is fed from a 2-pole (phases a-b), 480V.
- = angle of lead or angle of lag (radians) (current with respect to voltage in a single phase circuit) (subscript SP designates single phase) for a lagging power factor, θ SP < 0 for a leading power factor, θ SP > 0 2C. RMS Voltage and RMS Current A trace of instantaneous voltage as obtained with an oscilloscope is of interest and.
- In comparison to a single phase, three-phase power arrangements supply puts greater sums of power — 1.732 times the single phase — on the same current: In loads that use low amounts of power — such as the lights, television, radio, computer and scanner — the power can be supplied by any of the three 120 V single-phase power circuits

- Single-phase systems can be derived from three-phase systems. In the US, this is done via a transformer to get the proper voltage, while in the EU it is done directly. Voltage levels in the EU are such that a three-phase system can also serve as three single-phase systems. Single-phase vs. three-phase power. One other important difference.
- Since the load current is constant over the studied period, therefore the rms value of the load current I orms is The output AC power is given by: Problem 4.1 A single-phase full-wave uncontrolled bridge rectifier is supplying a highly inductive load (L/R ratio is very large), the load current is assumed to be smooth and ripple -free
- Single Phase Short-Circuit Current. Now, let´s assume the occurrence of single-phase (single-line-to-ground) short-circuit at point F. The value of short-circuit current is dependent on the zero-sequence connection of transformer T1 (which is given by the type of transformer and its winding connection).. Consider a shell-type transformer
- Hence all testing used in the IEEE 1584 empirical model was three-phase testing and, at this time, there is no single-phase model equation in the IEEE 1584. As a matter of fact, for single phase arc flash analysis, IEEE 1584 is endorsing a very conservative theoretical model developed by Ralph Lee [1] back in 80th for 3Φ systems
- This is a delta configuration . The phase current is current which flow in phase winding which is represented as IP here in above figure. Line current in current flowing in line which is represented as IL in the figure . Here each line current is.
- Summary: Single-phase induction motors. Single-phase induction motors are not self-starting without an auxiliary stator winding driven by an out of phase current of near 90°. Once started the auxiliary winding is optional. The auxiliary winding of a permanent split capacitor motor has a capacitor in series with it during starting and running

For DC and single-phase AC, the length times two is calculated because the current over + and - or L and N flows back and forth. The linking factor of three-phase current \(\mathbf{\sqrt{3}}\) is a fixed value. It arises from the interaction of the three phases, because the current does not just flow back and forth here A standard single-phase voltage or current source inverter can be in the half-bridge or full-bridge configuration. The single-phase units can be joined to have three-phase or multiphase topologies. Some industrial applications of inverters are for adjustable-speed ac drives, induction heating, standby aircraft power supplies, UP calculation of single phase, or phase-to-phase fault currents. The levels of fault current are decided by the armature reaction being created by the effect of the fault current flowing through the stator windings and the resulting effect of de-magnetising Ampere Turns on the rotor flux. The armature reaction to a 3 phase fault is a condition tha

The reactor must carry 12A (fundamental current) according to the NEC table for single-phase motor current. A 5% impedance is desired. For a 60Hz supply, the formula to calculate required inductance is: L = (ZV) / (377I), where L is inductance in Henries, Z is percent impedance, V is supply voltage, and I is full load amps Maximum AC Input Current 7843 VA / 187Vac = 41.94 amps (if it was single phase) Correction for Three Phase Input 41.94 amps / √3 (1.73) = 24.21 amps per phase So we have two equations, one for single phase inputs and one for three phase inputs: Single Phase Maximum Input Current Equatio The two legs of a single phase 240vac circuit, like the one found in most houses, are NOT out of phase with each other. They are in fact exactly IN phase with each other, meaning that there is only ONE phase. If the two legs were 180deg OUT of phase (technically two phases), measuring the voltage between the two legs would show 0v The single-line diagram of a three-phase system is shown in Fig. 1. Using the common base S b = 50 MVA, draw the impedance diagram in per unit including the load impedance. The manufacturer's nominal ratings are given as follows It is important to understand how the basic formula works even if the plan is to use software for the calculation. Below is a 3-step formula to calculate three-phase AFC, also called the available short circuit current (ISC) at the end of a run of wire: Step 1: F = (1.732 X L X I) ÷ (C X E_(L_L)) Step 2: Multiplier (M) = 1 ÷ (1 + F) Step 3

Due to the reduction of I2R, the current is low. Meaning, single-phase power ensures the unit to operate at optimum with an increased efficiency of its transmission. Single-phase power is best to use with fractional, or lower horsepower units up to 5 HP Transformer Formula. The following formulas are used to calculate the primary and secondary full-load currents. Single phase full load = tr * 1000 / v. three phase full load = tr * 1000/ (1.73 * V) Where tr is the transformer rating; v is the voltage (either primary or secondary) The following formula is used to calculate the transformer turns.

Use the single-phase equivalent circuit in each load configuration and calculate the phase currents for given values of the voltage and the load impedance. 5: Three incandescent lamps rated 60 W, 120 V (rms) are connected in the delta form 5.2 DIRECT CURRENT (DC) FORMULAS Basic Formulas Volts V = I x R Power in watts P V I P I R = × = 2 × 5.3 ALTERNATING CURRENT (AC) SINGLE PHASE V denotes line to neutral voltage. Basic Formulas Volts V =I ×Z Power Factor pf = cosθ Apparent Power = Reactive Power = VA V x I VARS V x I x sinθ Real Power W = V x I x p A single phase line consists of two parallel conductors which form a rectangular loop of one turn. When an alternating current flows through such a loop, a changing magnetic flux is set up. The changing flux links the loop and hence the loop (or single phase line) possesses inductance A three phase balanced system is analysed considering only one phase and neutral return. This is called per phase analysis. So the above analysis for single phase is true for balanced three phase case. Let the total power here is P t. Then we get total three phase power as thrice of single phase case As an example, consider a single-phase application of 2HP supplied by 240 Vac. The reactor must carry 12A (fundamental current) according to the NEC table for single-phase motor current. A 5% impedance is desired. For a 60Hz supply, the formula to calculate require