How to find the effective value of current. Active resistance. Effective values ​​of current and voltage – Knowledge Hypermarket

The physical meaning of these concepts is approximately the same as the physical meaning of average speed or other quantities averaged over time. At different times the strength AC and its voltage is accepted different meanings, therefore, talking about the strength of alternating current in general can only be conditional.

At the same time, it is quite obvious that different currents have different energy characteristics– they produce various jobs over the same period of time. The work produced by the current is taken as the basis for determining the effective value of the current. They are set for a certain period of time and calculate the work done by alternating current during this period of time. Then, knowing this work, they perform a reverse calculation: they find out the strength of the direct current that would produce similar work in the same period of time. That is, they perform averaging over power. The calculated force of a direct current hypothetically flowing through the same conductor, producing the same work, is the effective value of the original alternating current. The same applies to tension. This calculation comes down to determining the value of such an integral:

Where does this formula come from? From the well-known formula for current power, expressed through the square of its strength.

Effective values ​​of periodic and sinusoidal currents

Calculating the effective value for arbitrary currents is an unproductive task. But for periodic signal this parameter can be quite useful. It is known that any periodic signal can be decomposed into a spectrum. That is, represented as a finite or infinite sum of sinusoidal signals. Therefore, to determine the value of the effective value of such periodic current we need to know how to calculate the rms value of a simple sinusoidal current. As a result, by adding the effective values ​​of the first few harmonics with the maximum amplitude, we obtain an approximate value of the effective current value for an arbitrary periodic signal. Substituting the expression for harmonic vibration into the above formula, we obtain the following approximate formula.

Definition 1

Effective (effective) is the value of alternating current equal to the value of the equivalent direct current, which, when passing through the same resistance as alternating current, releases the same amount of heat over the same period of time.

Quantitative relationship between the amplitudes of AC force and voltage and effective values

The amount of heat released by alternating current at resistance $R$ over a short period of time $dt$ is equal to:

Then, in one period, alternating current releases heat ($W$):

Let us denote by $I_(ef)$ the strength of the direct current, which at the resistance $R$ releases the same amount of heat ($W$) as the alternating current $I$ in a time equal to the oscillation period of the alternating current ($T$). Then we express $W$ in terms of direct current and equate the expression to the right side of equation (2), we have:

Let us express from equation (3) the strength of the equivalent direct current, we obtain:

If the current varies according to a sinusoidal law:

Let's substitute expression (5) for alternating current into formula (4), then the magnitude of direct current will be expressed as:

Therefore, expression (6) can be transformed to the form:

where $I_(ef)$ is called the effective current value. Expressions for effective (effective) stress values ​​are written similarly:

Application of effective values ​​of current and voltage

When we talk about alternating current and voltage in electrical engineering, we mean their effective values. In particular, voltmeters and ammeters are usually calibrated to effective values. Hence, maximum value the voltage in the AC circuit is approximately 1.5 times greater than what the voltmeter shows. This fact should be taken into account when calculating insulators and studying safety problems.

The effective values ​​are used to characterize the alternating current (voltage) waveform. Thus, the amplitude coefficient ($k_a$) is introduced. equal:

and shape factor ($k_f$):

where $I_(sr\ v)=\frac(2)(\pi )\cdot I_m$ is the average rectified current value.

For sinusoidal current $k_a=\sqrt(2),\ k_f=\frac(\pi )(2\sqrt(2))=1.11.$

Example 1

Exercise: The voltage shown by the voltmeter is $U=220 V$. What is the voltage amplitude?

Solution:

As was said, voltmeters and ammeters are usually calibrated to effective voltage values ​​(current), therefore, the device shows in our notation $U_(ef)=220\V.$ In accordance with the known relationship:

Let's find the amplitude value of the voltage as:

Let's calculate:

Answer:$U_m\approx 310.2\ V.$

Example 2

Exercise: How is the alternating current power across resistance $R$ related to the effective values ​​of current and voltage?

Solution:

The average value of alternating current power in the circuit is

\[\left\langle P\right\rangle =\frac(A_T)(T)=\frac(U_mI_mcos\varphi )(2)\left(2.1\right),\]

where $cos\varphi$ is the power factor, which shows the efficiency of power transfer from the current source to the consumer. On the other hand, the average current powers at individual elements chains $\left\langle P_(tC)\right\rangle =0,\left\langle P_(tL)\right\rangle =0,\left\langle P_(tR)\right\rangle =\frac(1) (2)(I^2)_mR,$ and the resulting power can be found as the sum of powers:

\[\left\langle P\right\rangle =\left\langle P_(tC)\right\rangle +\left\langle P_(tL)\right\rangle +\left\langle P_(tR)\right\rangle \left(2.2\right).\]

Therefore, we can write that:

\[\left\langle P\right\rangle =P_(tR)=\frac(1)(2)(I^2)_mR=\frac(U_mI_mcos \varphi)(2)\left(2.3\right), \]

where $I_m\ $ is the amplitude of the current, $U_m$ is the amplitude of the external voltage, $\varphi$ is the phase difference between the current and voltage.

With direct current, the instantaneous power coincides with the average power. For $I_(ef)$=const we can set $cos\varphi =1,\ $which means formula (2.3) can be written as:

if instead of amplitude values ​​($U_m\ and\ I_m$) we use their effective (effective) values:

Therefore, the current power can be written as:

where $cos\varphi$ is the power factor. In technology, this coefficient is made as large as possible. When $cos\varphi $ is small, in order for the circuit to stand out required power need to skip high current, which leads to an increase in losses in the supply wires.

The same power (as in expression (2.3)) is developed by direct current, the strength of which is presented in formula (2.5).

Answer:$P_(tR)=U_(ef)I_(ef)cos\varphi .$

Values ​​of effective voltage and current. Definition. Relation to amplitude for different shapes. (10+)

The concept of effective (rms) values ​​of voltage and current

When we talk about variable voltage or current strength, especially of complex shape, then the question arises of how to measure them. After all, the tension is constantly changing. You can measure the amplitude of the signal, that is, the maximum modulus of the voltage value. This measurement method is fine for relatively smooth signals, but the presence of short bursts spoils the picture. Another criterion for choosing a measurement method is the purpose for which the measurement is being made. Since in most cases the interest is in the power that a particular signal can produce, the effective (effective) value is used.

The strength of alternating current (voltage) can be characterized using amplitude. However, the amplitude value of the current is not easy to measure experimentally. It is convenient to associate the strength of alternating current with any effect produced by the current, independent of its direction. This is, for example, the thermal effect of current. The rotation of the needle of an ammeter that measures alternating current is caused by the elongation of the filament, which heats up when current passes through it.

Current or effective The value of alternating current (voltage) is the value of direct current at which the same amount of heat is released at the active resistance over a period as with alternating current.

Let us connect the effective value of the current with its amplitude value. To do this, let's calculate the amount of heat generated at the active resistance by alternating current in a time equal to the oscillation period. Let us recall that according to the Joule-Lenz law, the amount of heat released in a section of the circuit with resistance permanent current in time , is determined by the formula
. Alternating current can be considered constant only for very short periods of time
. Let's divide the oscillation period for a very large number of small periods of time
. Amount of heat
, allocated at the resistance in time
:
. The total amount of heat released over a period can be found by summing up the heat released over individual short periods of time, or, in other words, by integrating:

.

The current strength in the circuit varies according to a sinusoidal law

,

.

Omitting the calculations associated with integration, we write the final result

.

If some direct current were flowing through the circuit , then in a time equal to , heat would be released
. By definition, direct current , which has the same thermal effect as alternating current, will be equal to the effective value of alternating current
. We find the effective value of the current by equating the heat released over a period in the cases of direct and alternating currents

(4.28)

Obviously, exactly the same relationship connects the effective and amplitude values ​​of the voltage in a circuit with a sinusoidal alternating current:

(4.29)

For example, the standard network voltage of 220 V is the effective voltage. Using formula (4.29), it is easy to calculate that the amplitude value of the voltage in this case will be equal to 311 V.

4.4.5. AC power

Let in some section of the circuit with alternating current the phase shift between current and voltage be equal to , i.e. Current and voltage change according to the laws:

,
.

Then the instantaneous value of the power released in the circuit section is

Power changes over time. Therefore, we can only talk about its average value. Let's define average power, released over a fairly long period of time (many times longer than the oscillation period):

Using the well-known trigonometric formula

.

Size
there is no need to average, since it does not depend on time, therefore:

.

Over a long period of time, the value of the cosine manages to change many times, taking on both negative and positive values ​​ranging from (1) to 1. It is clear that the average value of the cosine over time is zero

, That's why
(4.30)

Expressing the amplitudes of current and voltage through their effective values ​​using formulas (4.28) and (4.29), we obtain

. (4.31)

The power released in the AC section of the circuit depends on the effective values ​​of current and voltage and phase shift between current and voltage. For example, if a section of a circuit consists of only active resistance, then
And
. If a section of a circuit contains only inductance or only capacitance, then
And
.

The average zero value of power allocated to inductance and capacitance can be explained as follows. Inductance and capacitance only borrow energy from the generator and then return it back. The capacitor charges and then discharges. The current strength in the coil increases, then drops again to zero, etc. It is for the reason that the average energy consumed by the generator at inductive and capacitive reactances is zero, they were called reactive. At active resistance, the average power is different from zero. In other words, a wire with resistance When current flows through it, it heats up. And the energy released in the form of heat does not return back to the generator.

If a section of the circuit contains several elements, then the phase shift may be different. For example, in the case of the circuit section shown in Fig. 4.5, the phase shift between current and voltage is determined by formula (4.27).

Example 4.7. A resistor with resistance is connected to the alternating sinusoidal current generator . How many times will the average power consumed by the generator change if a coil with inductive reactance is connected to a resistor?
a) in series, b) in parallel (Fig. 4.10)? Neglect the active resistance of the coil.

Solution. When only active resistance is connected to the generator , power consumption

(see formula (4.30)).

Consider the circuit in Fig. 4.10, a. In example 4.6, the amplitude value of the generator current was determined:
. From the vector diagram in Fig. 4.11,a we determine the phase shift between the current and voltage of the generator

.

As a result, the average power consumed by the generator

.

Answer: when connected in series to an inductance circuit, the average power consumed by the generator will decrease by 2 times.

Consider the circuit in Fig. 4.10, b. In example 4.6, the amplitude value of the generator current was determined
. From the vector diagram in Fig. 4.11b we determine the phase shift between the current and voltage of the generator

.

Then the average power consumed by the generator

Answer: when inductance is connected in parallel, the average power consumed by the generator does not change.

When calculating alternating current circuits, they usually use the concept of effective (effective) values ​​of alternating current, voltage and e. d.s.

Effective values ​​of current, voltage and e. d.s. are indicated by capital letters.

On the scales of measuring instruments and technical documentation The actual values ​​of the quantities are also indicated.

The effective value of the alternating current is equal to the value of the equivalent direct current, which, passing through the same resistance as the alternating current, releases the same amount of heat over a period.

The amount of heat released by alternating current in resistance in an infinitesimal period of time

and for the period of alternating current T

Equating the resulting expression to the amount of heat released in the same resistance DC for the same time T, we get:

By reducing common multiplier, we obtain the effective current value

Rice. 5-8. Graph of alternating current and current squared.

In Fig. 5-8, a curve of instantaneous values ​​of current i and a curve of squared instantaneous values ​​are plotted. The area bounded by the last curve and the abscissa axis is, on a certain scale, a value determined by the expression The height of a rectangle equal to the area bounded by the curve and the abscissa axis, equal to the average value of the ordinates of the curve, is square of effective current value

If the current changes according to the sine law, i.e.

Similarly for the effective values ​​of sinusoidal voltages and e. d.s. you can write:

In addition to the effective value of current and voltage, sometimes they also use the concept of the average value of current and voltage.

The average value of the sinusoidal current over a period is zero, since during the first half of the period a certain amount of electricity Q passes through the cross-section of the conductor in the forward direction. During the second half of the period, the same amount of electricity passes through the cross-section of the conductor in the opposite direction. Consequently, the amount of electricity passing through the cross-section of the conductor during a period is equal to zero, and the average value of the sinusoidal current over the period is also equal to zero.

Therefore, the average value of the sinusoidal current is calculated over the half-cycle during which the current remains positive. The average value of the current is equal to the ratio of the amount of electricity passing through the cross-section of the conductor in half a period to the duration of this half-cycle.