How to find out the current strength, knowing the power and voltage.
To answer the question of how to determine current, you need to divide the electrical voltage by the total number of watts. In this case, you can do all the necessary calculations yourself, or you can resort to a special online calculator.
Calculation of power indicator by amperes and watts.
You can find out the electricity consumption by the current strength of the resistor by multiplying the former by the resistance, expressed in Ohms. The result is a value expressed in volts multiplied by ohms. It turns out ampere.
Note! If there is no resistance, you need to divide the watt indicator by the current energy, that is, you should divide the watts by amperes and you will get the value of electricity in volts. You can understand the power reading through the amount of electricity with electrical voltage by multiplying the corresponding readings from the device.
Calculation of electricity through electrical power and voltage
What is the current in the outlet - direct or alternating?
People who are more or less familiar with electrical engineering can easily answer the question of what current is in the outlet. Of course it's variable. This type of electricity is much easier to produce and transmit over long distances, and therefore the choice in favor of alternating current is obvious.
There are two types of current - direct and alternating. To understand the difference and determine whether the outlet has direct or alternating current, you should delve into some technical features. Alternating current has the property of changing in direction and magnitude. Direct current has stable qualities and direction of movement of charged particles.
Alternating current comes out of the power plant generators with a voltage of 220-440 thousand volts. When approaching an apartment building, the current is reduced to 12 thousand volts, and at the transformer station it is converted to 380 volts.
Advice
The voltage between phases is called linear. The low-voltage section of the step-down substation produces three phases and a zero (neutral) wire. Energy consumers are connected from one of the phases and the neutral wire.
Thus, single-phase alternating current with a voltage of 220 volts enters the building.
The distribution diagram of electricity between houses is presented below:
In the home, electricity is supplied to the meter, and then through automatic machines to the boxes of each room. The boxes contain wiring throughout the room for a couple of circuits - electrical outlets and lighting equipment.
The machines can be provided one for each room or one for each circuit.
Taking into account how many amperes the outlet is designed for, it can be included in a group or connected to a dedicated circuit breaker.
Alternating current accounts for approximately 90% of all electricity consumed. Such a high specific gravity is due to the peculiarities of this type of current - it can be transported over considerable distances by changing the voltage at substations to the required parameters.
Sources of direct current are most often batteries, galvanic cells, solar panels, thermocouples.
Direct current is widely used in local networks of automobile and air transport, in computer electrical circuits, automatic systems, radio and television equipment.
Direct current is used in contact networks of railway transport, as well as on ship installations.
The diagram below shows the fundamental differences between direct and alternating currents.
Home electrical network parameters
The main parameters of electricity are its voltage and frequency. The standard voltage for home electrical networks is 220 volts. The generally accepted frequency is 50 hertz. However, in the USA a different frequency value is used - 60 hertz. The frequency parameter is set by the generating equipment and is unchanged.
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The voltage in the network of a particular house or apartment may be different from the nominal value (220 volts). This indicator is influenced by the technical condition of the equipment, network loads, and substation load. As a result, the voltage may deviate from the specified parameter in one direction or another by 20–25 volts.
Current load
All sockets have a certain marking, by which you can judge the permissible current load. For example, the designation "5A" indicates a maximum current of 5 amperes. Acceptable indicators must be observed, since otherwise the equipment may fail, including fire.
The markings on the sockets are shown in the figure below:
All legally sold electrical appliances are accompanied by a passport indicating the power consumption or current load rating.
The largest consumers of electricity are household appliances such as air conditioners, microwave ovens, washing machines, electric stoves and ovens.
For normal operation, such devices will need an outlet with a load of at least 16 amperes.
note
If the documentation for electrical household appliances does not contain information about the consumed amperes (current strength in the outlet), the required values are determined using the electric power formula:
The power indicator is in the passport, the network voltage is known. To determine electricity consumption, you need to divide the power indicator (indicated only in watts) by the voltage value.
Calculation of power consumption
Electric power is a quantity that is responsible for the fact of the rate of change or transmission of electrical energy. There is full and active power load, as well as active and reactive. The total is calculated as follows: S = √ (P2 + Q2), where P is the active part and Q is the reactive part. To find the consumed power indicator, you need to know the amount of electric current that is consumed by the load, as well as the supply voltage that is supplied by the source.
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As for the household determination of consumed electrical energy, it is necessary to calculate the total number of watts of power supply for electrical appliances and the passport data of the rated electric current of the boiler. As a rule, all electrical appliances operate with alternating current and a voltage of 220 volts. The easiest way to calculate current is to use an ammeter. Knowing the first and second parameters, it is possible to find out the amount of energy consumed.
It is worth pointing out that measuring power through voltage or calculating power from resistance and voltage is possible not only with a formula, but also with a device. To do this, you can use a multimeter with a current clamp or a specialized meter - a wattmeter.
Note! Both work on the same principle, as stated in their instruction manual.
Power consumption calculation
Power, current and voltage are the three components of calculating wiring in a house. Find out all the necessary parameters in any network simply using the formulas presented above. The proper operation of all home electrical equipment and the safety of its owner will depend on these values.
What affects current power
Adding electrical resistance allows you to take into account losses in the connected circuit (load). The formula for finding power for a complete circuit takes into account the parameters of the power source. For a more accurate analysis, the rate of energy consumption per unit volume of conductor (ΔV) should be estimated.
Power is equal to the formula:
Psp = Rsp * j2,
Where:
- Rsp – resistivity;
- j is the current density of the corresponding section of the circuit.
From this expression the dependence of electricity consumption on conductivity is clear. This ratio determines the requirements for the cable products used. If the cross-section is insufficient (high level of impurities), heating increases. A similar result is obtained when connecting a powerful load. At a certain level, thermal destruction of the material will occur.
For your information. This process is the cause of typical emergency situations. To prevent damage, specialized equipment is used - circuit breakers.
Differences in power at constant and alternating voltage
We use the designations of electrical quantities that are accepted in our country:
- P – active power, measured in watts, denoted by W;
- Q – reactive power, measured in reactive volt amperes, denoted by VAR;
- S – total power, measured in volt amperes, designated VA;
- U – voltage, measured in volts, denoted by VA;
- I – current, measured in amperes, denoted by A;
- R - resistance, measured in ohms, denoted by Ohm.
Let us name the main differences between P on a constant current and Q on an alternating electric current. Calculating P for a constant electric current is the simplest. For sections of an electrical circuit, Ohm's law is valid. This law involves only the magnitude of the applied U (voltage) and the magnitude of the resistance R.
Calculation of S (total power) on alternating electric current is somewhat more complicated. In addition to P, there is Q and the concept of power factor is introduced. Algebraically adding the active P and reactive Q, we get the total S.
AC voltage
As we know from physics lessons, current is the movement of charged particles that occurs under the influence of an electromagnetic field, potential difference and tension. The main characteristic of any voltage is its dependence on time. Based on this, a distinction is made between constant and variable quantities. The value of a constant practically does not change over time, but the value of a variable changes.
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Ohm's law
In turn, a variable characteristic can be periodic or non-periodic. Periodic is a voltage whose values are repeated at regular intervals. The non-periodic is capable of changing at any period of time.
Scheme for describing the physical meaning
Voltage in an alternating circuit is a parameter that changes its value over time. To simplify explanations, sinusoidal harmonic alternating voltage will be considered in the following.
The minimum time during which a variable repeats is called a period. Absolutely any periodic quantity can be written as a dependence on any function. If time is t, then the dependence will be denoted by F(t). Thus, any period in time has the form: F(t+-T) = F(t), where T is the period.
The physical quantity that is the reciprocal of the period is called frequency. It is equal to 1/T. Its unit of measurement is the hertz, while the unit of measurement of the period is the second.
f = 1/T, 1 Hz = 1/s = s to the minus first power.
Oscillation formulas
Important! The most common functional dependence of a variable network is in the form of a sinusoid. That is why it was taken as the basis for this material.
It is known from mathematics that a sinusoid is the simplest periodic function, and with its help, any other periodic functions can be represented from several sinusoids with multiple frequencies.
Sinusoidal voltage in absolutely any period of time can be described by the instantaneous characteristic: u = U * sin(ωt + φ), where ω = 2πf = 2π/T, where U is the maximum voltage (amplitude), ω is the angular rate of change, φ is the initial phase, which is determined by the displacement of the function relative to the zero coordinate point.
Sine function
Part (ωt + φ) is a phase that characterizes the voltage value in a specific period of time. From this it turns out that amplitude, angular velocity and phase are the main characteristics of variable networks that determine their values in any time interval.
Important! When considering a sine function, the phase is often taken to be zero. In practice, they also often resort to some other parameters, including effective and average voltage, shape factor.
AC voltage regulator
Types of capacities
Power is a measured physical quantity that is equal to the rate of change with the conversion, transmission or consumption of system energy. According to a narrower concept, this is an indicator that is equal to the ratio of time spent on work to the period itself spent on work. In mechanics it is designated by the symbol N. In electrical engineering the letter P is used. You can often also see the symbol W, from the word watt.
AC power is the product of current, voltage and cosine of the phase shift. In this case, only the active and reactive varieties can be easily counted. You can find out the full power value through the vector dependence of these indicators and area.
Main power types.
Current power through resistor
Let alternating current flow through a resistor with a resistance of . The voltage across the resistor, as we know, oscillates in phase with the current:
Therefore, for instantaneous power we obtain:
(2)
We see that the power is non-negative all the time - the resistor takes energy from the circuit, but does not return it back to the circuit.
AC power through a resistor.
The maximum value of our power is related to the amplitudes of current and voltage by the usual formulas:
In practice, however, it is not the maximum, but the average current power that is of interest. This is understandable. Take, for example, an ordinary light bulb that is lit in your home. A current with a frequency of Hz flows through it, i.e., fluctuations in current and voltage occur per second. It is clear that over a sufficiently long time, a certain average power is released from the light bulb, the value of which is somewhere between and . Where exactly?
Look again carefully at Fig. 1. Don’t you have an intuitive feeling that the average power corresponds to the “middle” of our sinusoid and therefore takes on the value ?
This feeling is absolutely true! The way it is. Of course, we can give a mathematically strict definition of the average value of a function (in the form of some integral) and confirm our guess with direct calculation, but we don’t need this. An intuitive understanding of a simple and important fact is enough:
the average value of the square of the sine (or cosine) over the period is equal to .
The mean value of the squared sine is
So, for the average value of the current power across the resistor we have:
(3)
In connection with these formulas, the so-called effective (or effective) values of voltage and current are introduced (in fact, these are nothing more than the root mean square values of voltage and current. We have already encountered this: the root mean square speed of the molecules of an ideal gas (leaf "Equation of state of an ideal gas"):
(4)
Formulas (3), written in terms of effective values, are completely similar to the corresponding formulas for direct current:
Therefore, if you take a light bulb, connect it first to a constant voltage source, and then to an alternating voltage source with the same effective value, then in both cases the light bulb will burn equally brightly.
The effective values (4) are extremely important for practice. It turns out that AC voltmeters and ammeters show exactly the effective values (that’s how they are designed). Know also that the notorious volts from the outlet are the effective value of the household power supply voltage.
Current power through the capacitor
Let an alternating voltage be applied to the capacitor. As we know, the current through a capacitor is ahead of the voltage in phase by:
For instantaneous power we get:
Graph of instantaneous power versus time.
AC power through a capacitor.
What is the average power value? It corresponds to the “middle” of the sinusoid and in this case is equal to zero! We see this now as a mathematical fact. But it would be interesting to understand from a physical point of view why the current power through the capacitor turns out to be zero.
To do this, let's draw graphs of the voltage and current in the capacitor over one oscillation period.
The voltage across the capacitor and the current through it.
Let's consider all four quarters of the period sequentially.
1. First quarter, . The tension is positive and increasing. The current is positive (flows in the positive direction), the capacitor is charged. As the charge on the capacitor increases, the current decreases.
Instantaneous power is positive: the capacitor stores energy coming from the external circuit. This energy arises due to the work of an external electric field that pushes charges onto the capacitor.
2. Second quarter, . The tension remains positive, but is subsiding. The current changes direction and becomes negative: the capacitor is discharged against the direction of the external electric field. At the end of the second quarter, the capacitor is completely discharged.
Instantaneous power is negative: the capacitor releases energy. This energy returns to the circuit: it is used to do work against the electric field of the external circuit (the capacitor, as it were, “pushes” the charges in the direction opposite to the one in which the external field “wants” to move them).
3. Third quarter, . The external electric field changes direction: the voltage is negative and increases in magnitude. The current is negative: the capacitor is charging in the negative direction.
The situation is completely similar to the first quarter, only the signs of voltage and current are opposite. The power is positive: the capacitor again stores energy.
4. Fourth quarter, . The voltage is negative and decreases in magnitude. The capacitor is discharged against the external field: the current strength is positive.
Power is negative: the capacitor returns energy to the circuit. The situation is similar to the second quarter - again with the replacement by replacing the signs of current and voltage with opposite ones.
We see that the energy taken by the capacitor from the external circuit during the first quarter of the oscillation period is completely returned to the circuit during the second quarter. This process is then repeated again and again. This is why the average power consumed by a capacitor turns out to be zero.
Voltage formula
There is a formula in physics, although it has no practical application. The official formula is written like this.
voltage formula
Where
A is the work done by the electric field to move a charge along a section of the circuit, Joules
q - charge, Coulomb
U—voltage on a section of the electrical circuit, Volts
In practice, the voltage across a section of a circuit is derived through Ohm's law.
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voltage from Ohm's law
Where
I - current, Amperes
R - resistance, Ohms
Current power through the coil
Let alternating voltage be applied to the coil. The current through the coil lags in phase from the voltage by:
For instantaneous power we get:
Again the average power is zero. The reasons for this are, in general, the same as in the case of a capacitor. Let's look at the graphs of voltage and current through the coil over a period (Fig. 5).
The voltage across the coil and the current through it.
We see that during the second and fourth quarters of the period, energy enters the coil from the external circuit. In fact, voltage and current have the same signs, the current increases in magnitude; To create a current, the external electric field does work against the vortex electric field, and this work goes to increase the energy of the magnetic field of the coil.
In the first and third quarters of the period, the voltage and current have different signs: the coil returns energy to the circuit. The vortex electric field, which maintains a decreasing current, moves charges against the external electric field and thereby does positive work. How is this work accomplished? Due to the energy previously accumulated in the coil.
Thus, the energy stored in the coil during one quarter of the period is completely returned to the circuit during the next quarter. Therefore, the average power consumed by the coil is zero.
