Lr circuit calculator

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RL series circuit

Formulas and description for RL in series


The total resistance of the RL series in the AC circuit is referred to as the impedance Z. Ohm's law applies to the entire circuit.

The current is the same at every measuring point. Current and voltage are in phase at the ohmic resistance. In the inductive reactance of the coil the current lag the voltage by −90 °.

The total voltage U is the sum of the geometrically added partial voltages. For this purpose, both partial voltages form the legs of a right triangle. Its hypotenuse corresponds to the total voltage U. The resulting triangle is called the voltage triangle or vector diagram of the voltages.

\(\displaystyle U\) Applied voltage
\(\displaystyle U_L\) Voltage on the coil
\(\displaystyle U_R\) Voltage across the resistor
\(\displaystyle I\) Current
\(\displaystyle R\) Ohmic resistance
\(\displaystyle X_L\) Inductive reactance
\(\displaystyle Z\) Impedance
\(\displaystyle S\) Apparent power
\(\displaystyle Q\) Inductive reactive power
\(\displaystyle p\) Real power
\(\displaystyle φ\) Phase shift in °

Voltage triangle

\(\displaystyle U=\sqrt{ {U_R}^2 + {U_L}^2} \)
\(\displaystyle U_R=\sqrt{U^2-{U_L}^2} \) \(\displaystyle =U·cos(φ) \)
\(\displaystyle U_L=\sqrt{U^2 - {U_R}^2} \) \(\displaystyle =U · sin(φ) \)
\(\displaystyle φ =arctan\left( \frac{X_L}{R} \right) \)

Resistance triangle

\(\displaystyle Z=\sqrt{R^2 + {X_L}^2} \) \(\displaystyle =\frac{U}{I} \)
\(\displaystyle R=\sqrt{Z^2-{X_L}^2} \) \(\displaystyle =\frac{U_R}{I} \) \(\displaystyle =Z·cos(φ) \)
\(\displaystyle X_L=\sqrt{Z^2-R^2} \) \(\displaystyle =\frac{U_L}{I} \) \(\displaystyle =Z·sin(φ) \)
\(\displaystyle φ =arctan\left( \frac{U_L}{U_R} \right) \)

Power triangle

\(\displaystyle S=\sqrt{P^2 + Q^2} \) \(\displaystyle =U·I \)
\(\displaystyle P=\sqrt{S^2-Q^2} \) \(\displaystyle =U_R· I \) \(\displaystyle =S·cos(φ) \)
\(\displaystyle Q=\sqrt{S^2-P^2} \) \(\displaystyle = U_L· I\) \(\displaystyle =S·sin(φ) \)

Power factor

\(\displaystyle cos(φ)=\frac{P}{s}=\frac{U_R}{U}=\frac{R}{Z}\)
Sours: https://www.redcrab-software.com/en/Calculator/Electrics/RL-Series-Circuit

Learning Objectives

By the end of this section, you will be able to:

  • Calculate the current in an RL circuit after a specified number of characteristic time steps.
  • Calculate the characteristic time of an RL circuit.
  • Sketch the current in an RL circuit over time.

We know that the current through an inductor L cannot be turned on or off instantaneously. The change in current changes flux, inducing an emf opposing the change (Lenz’s law). How long does the opposition last? Current will flow and can be turned off, but how long does it take? Figure 1 shows a switching circuit that can be used to examine current through an inductor as a function of time.

Part a of the figure shows an inductor connected in series with a resistor. The arrangement is connected across a cell by an on and off switch with two positions. When in position one, the battery, resistor, and inductor are in series and a current is established. In position two, the battery is removed and the current stops eventually because of energy loss in the resistor. Part b of the diagram shows the graph when the switch is in position one. It shows a graph for current growth verses time. The current is along the Y axis and the time is along the X axis. The graph shows a smooth rise from origin to a maximum value I zero corresponding to Y axis and value four tau on X axis. Part c of the diagram shows the graph when the switch is in position two. It shows a graph for current decay verses time is shown. The current is along the Y axis and the time is along the X axis. The graph is decreasing curve from a value I zero on Y axis, touching the X axis at a point where value of time equals four tau.

Figure 1. (a) An RL circuit with a switch to turn current on and off. When in position 1, the battery, resistor, and inductor are in series and a current is established. In position 2, the battery is removed and the current eventually stops because of energy loss in the resistor. (b) A graph of current growth versus time when the switch is moved to position 1. (c) A graph of current decay when the switch is moved to position 2.

When the switch is first moved to position 1 (at = 0), the current is zero and it eventually rises to IV/R, where R is the total resistance of the circuit. The opposition of the inductor L is greatest at the beginning, because the amount of change is greatest. The opposition it poses is in the form of an induced emf, which decreases to zero as the current approaches its final value. The opposing emf is proportional to the amount of change left. This is the hallmark of an exponential behavior, and it can be shown with calculus that

I(1 − et/τ)       (turning on),

is the current in an RL circuit when switched on (Note the similarity to the exponential behavior of the voltage on a charging capacitor). The initial current is zero and approaches IV/R with a characteristic time constantτ for an RL circuit, given by

[latex]\tau =\frac{L}{R}\\[/latex],

where τ has units of seconds, since 1 H = 1 Ω·s. In the first period of time τ, the current rises from zero to 0.632 I0, since I0(1 − e−1) = I(1 − 0.368) = 0.632 I0. The current will go 0.632 of the remainder in the next time τ. A well-known property of the exponential is that the final value is never exactly reached, but 0.632 of the remainder to that value is achieved in every characteristic time τ. In just a few multiples of the time τ, the final value is very nearly achieved, as the graph in Figure 1(b) illustrates.

The characteristic time τ depends on only two factors, the inductance L and the resistance R. The greater the inductance L, the greater τ is, which makes sense since a large inductance is very effective in opposing change. The smaller the resistance R, the greater τ is. Again this makes sense, since a small resistance means a large final current and a greater change to get there. In both cases—large L and small R —more energy is stored in the inductor and more time is required to get it in and out.

When the switch in Figure 1(a) is moved to position 2 and cuts the battery out of the circuit, the current drops because of energy dissipation by the resistor. But this is also not instantaneous, since the inductor opposes the decrease in current by inducing an emf in the same direction as the battery that drove the current. Furthermore, there is a certain amount of energy,[latex]\left(\text{1/2}\right){{LI}_{0}}^{2}\\[/latex], stored in the inductor, and it is dissipated at a finite rate. As the current approaches zero, the rate of decrease slows, since the energy dissipation rate is I2R. Once again the behavior is exponential, and I is found to be

I0et/τ          (turning off).

(See Figure 1(c).) In the first period of time τ L/R after the switch is closed, the current falls to 0.368 of its initial value, since I0e−1 = 0.368 I0. In each successive time τ, the current falls to 0.368 of the preceding value, and in a few multiples of τ, the current becomes very close to zero, as seen in the graph in Figure 1(c).

Example 1. Calculating Characteristic Time and Current in an RL Circuit

(a) What is the characteristic time constant for a 7.50 mH inductor in series with a 3.00 Ω resistor? (b) Find the current 5.00 ms after the switch is moved to position 2 to disconnect the battery, if it is initially 10.0 A.

Strategy for (a)

The time constant for an RL circuit is defined by τ L/R.

Solution for (a)

Entering known values into the expression for τ given in τ L/R yields

[latex]\tau =\frac{L}{R}=\frac{7.50 \text{ mH}}{3.00\text{ }\Omega}=2.50 \text{ ms}\\[/latex].

Discussion for (a)

This is a small but definitely finite time. The coil will be very close to its full current in about ten time constants, or about 25 ms.

Strategy for (b)

We can find the current by using I0et/τ, or by considering the decline in steps. Since the time is twice the characteristic time, we consider the process in steps.

Solution for (b)

In the first 2.50 ms, the current declines to 0.368 of its initial value, which is

[latex]\begin{array}{lll}I& =& 0.368\text{ }I_{0}=\left(0.368\right)\left(10.0 \text{ A}\right)\\ & =& 3.68 \text{ A at }t=2.50\text{ ms}\end{array}\\[/latex].

After another 2.50 ms, or a total of 5.00 ms, the current declines to 0.368 of the value just found. That is,

[latex]\begin{array}{lll}I'& =& 0.368\text{ }I=\left(0.368\right)\left(3.68 \text{ A}\right)\\ & =& 1.35\text{ A at }t=5.00\text{ ms}\end{array}\\[/latex].

Discussion for (b)

After another 5.00 ms has passed, the current will be 0.183 A (see Exercise 1 in the Problems & Exercises section); so, although it does die out, the current certainly does not go to zero instantaneously.

In summary, when the voltage applied to an inductor is changed, the current also changes, but the change in current lags the change in voltage in an RL circuit. In Reactance, Inductive and Capacitive, we explore how an RL circuit behaves when a sinusoidal AC voltage is applied.

Section Summary

  • When a series connection of a resistor and an inductor—an RL circuit—is connected to a voltage source, the time variation of the current is

    I(1 − et/τ)       (turning on),

    where I0 = V/R is the final current.

  • The characteristic time constant τ is [latex]\tau =\frac{L}{R}\\[/latex], where L is the inductance and R is the resistance.
  • In the first time constant τ, the current rises from zero to 0.632I0, and 0.632 of the remainder in every subsequent time interval τ.
  • When the inductor is shorted through a resistor, current decreases as

    I0et/τ          (turning off).

    Here I0 is the initial current.

  • Current falls to 0.368I0 in the first time interval τ, and 0.368 of the remainder toward zero in each subsequent time τ.

Problems & Exercises

1. If you want a characteristic RL time constant of 1.00 s, and you have a 500 Ω resistor, what value of self-inductance is needed?

2. Your RL circuit has a characteristic time constant of 20.0 ns, and a resistance of 5.00 MΩ. (a) What is the inductance of the circuit? (b) What resistance would give you a 1.00 ns time constant, perhaps needed for quick response in an oscilloscope?

3. A large superconducting magnet, used for magnetic resonance imaging, has a 50.0 H inductance. If you want current through it to be adjustable with a 1.00 s characteristic time constant, what is the minimum resistance of system?

4. Verify that after a time of 10.0 ms, the current for the situation considered in Example 1 (above) will be 0.183 A as stated.

5. Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and resistors ranging from 0.100 Ω to 1.00 MΩ. What is the range of characteristic RL time constants you can produce by connecting a single resistor to a single inductor?

6. (a) What is the characteristic time constant of a 25.0 mH inductor that has a resistance of 4.00 Ω? (b) If it is connected to a 12.0 V battery, what is the current after 12.5 ms?

7. What percentage of the final current Iflows through an inductor L in series with a resistor R, three time constants after the circuit is completed?

8. The 5.00 A current through a 1.50 H inductor is dissipated by a 2.00 Ω resistor in a circuit like that in Figure 1 (above) with the switch in position 2. (a) What is the initial energy in the inductor? (b) How long will it take the current to decline to 5.00% of its initial value? (c) Calculate the average power dissipated, and compare it with the initial power dissipated by the resistor.

9. (a) Use the exact exponential treatment to find how much time is required to bring the current through an 80.0 mH inductor in series with a 15.0 Ω resistor to 99.0% of its final value, starting from zero. (b) Compare your answer to the approximate treatment using integral numbers of τ. (c) Discuss how significant the difference is.

10. (a) Using the exact exponential treatment, find the time required for the current through a 2.00 H inductor in series with a 0.500 Ω resistor to be reduced to 0.100% of its original value. (b) Compare your answer to the approximate treatment using integral numbers of τ. (c) Discuss how significant the difference is.

Glossary

characteristic time constant:
denoted by τ, of a particular series RL circuit is calculated by [latex]\tau =\frac{L}{R}\\[/latex], where is the inductance and R is the resistance

Selected Solutions to Problems & Exercises

1. 500 H

3. 50.0 Ω

5. 1.00 × 10–18 s to 0.100 s

7. 95.0%

9. (a) 24.6 ms (b) 26.7 ms (c) 9% difference, which is greater than the inherent uncertainty in the given parameters.

Sours: https://courses.lumenlearning.com/physics/chapter/23-10-rl-circuits/
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Scheme

This parallel RL circuit impedance calculator determines the impedance and the phase difference angle of an inductor and a resistor connected in parallel for a given frequency of a sinusoidal signal. The angular frequency is also determined.

Example:Calculate the impedance of a 500 mH inductor and a 200 kΩ resistor at a frequency of 25 kHz.

Input

Resistance, R

Inductance, L

Frequency, f

Share

Output

Angular Frequencyω= rad/s

Inductive reactanceXL= Ω

Total RL impedance|ZRL|= Ω

Phase differenceφ = ° = rad

Enter the resistance, inductance, and frequency values, select the units and click or tap the Calculate button. Try to enter zero or infinitely large values to see how this circuit behaves. Infinite frequency is not supported. To enter the Infinity value, just type inf in the input box.

The following formulas are used for the calculation:

Formula

Formula

Formula

Formula

Formula

Formula

Formula

where

The phasor diagram for a parallel RL circuit shows that the total current wave lags behind the total voltage wave. The lag is less than 90° and more than 0°. At 90° the resistor is removed from the circuit (the circuit is purely inductive) and at 0° the inductor is removed from the circuit (the circuit is purely resistive)

The phasor diagram for a parallel RL circuit shows that the total current wave lags behind the total voltage wave. The lag is less than 90° and more than 0°. At 90° the resistor is removed from the circuit (the circuit is purely inductive) and at 0° the inductor is removed from the circuit (the circuit is purely resistive)

ZRL is the RL circuit impedance in ohms (Ω),

ω = 2πf is the angular frequency in rad/s,

f is the frequency in hertz (Hz),

R is the resistance in ohms (Ω),

L is the inductance in henries (H),

φ is the phase shift between the total voltage VT and the total current IT in degress (°) and radians, and

j is the imaginary unit.

To calculate, enter the inductance, the resistance, and the frequency, select the units of measurements and the result for RL impedance will be shown in ohms and for the phase difference in degrees and radians. The inductive reactance in ohms will also be calculated.

Picture

A graph of the parallel RL circuit impedance ZRL against frequency f for a given inductance and resistance

For the parallel RL circuit, the impedance is a complex number and is determined as

Formula

Formula

Formula

The applied voltage VT is the same across both the resistor and the inductor. The total current IT is divided into the two branch currents IL and IR:

Formula

Formula

From the Kirchhoff's current law, the total current IT is the phasor sum of the two branch currents IL and IR, which are 90° out of phase with each other. Therefore,

Formula

Formula

Formula

Failure Modes

What if something goes wrong in this circuit? Click or tap a corresponding link to view the calculator in various failure modes:

Special Modes

Click or tap a corresponding link to view the calculator in various special modes:

Various direct current modes

Short circuit

Open circuit

Purely inductive circuit

Inductive circuit

Notes

  • In our explanations of the behavior of this circuit, zero frequency means direct current. If f = 0, we assume that the circuit is connected to an ideal DC voltage source.
  • At zero frequency, we consider the reactance of an ideal inductor to be infinitely large if its inductance is infinitely large. If the inductor has a finite inductance, its reactance at zero frequency is zero and for a DC voltage source, it represents a short circuit.

This article was written by Anatoly Zolotkov

Sours: https://www.translatorscafe.com/unit-converter/en-US/calculator/parallel-rl-impedance/
RL Circuits - Inductors \u0026 Resistors

A 200 Ω resistor and a 50 Ω XL are placed in series with a voltage source, and the total current flow is 2 amps, as shown in Figure.

Calculate Power in Series RL Circuit

Figure : Series R-L Circuit

Find:

  1. Power Factor, pf
  2. Applied voltage, V
  3. True Power, P
  4. Reactive Power, Q
  5. Apparent Power, S

Solution:

1. Calculate Power factor (pf)

Calculate Power Factor

p.f. = cos (14º)

p.f. = 0.097

2. Calculate Applied Voltage, V

V = I Z

Z = √R2 + XL2

so, V = I√R2 + XL2

V = 2√2002 + 502

V = 2√42500

V = 2 x 206.16

V = 412.3 Volts

3. Calculate True Power, P

P = EI cos θ

P = (412.3)(2)(0.97)

P = 799.86 watts

4. Calculate Reactive Power, Q

Q = EI sin θ

Q = (412.3)(2)(0.242)

Q = 199.6 VAR

5. Calculate Apparent Power, S

S = EI

S = (412.3)(2)

S = 824.6 VA

Categories Electrical TheorySours: https://instrumentationtools.com/calculate-power-in-series-rl-circuit/

Calculator lr circuit

Series RL circuit Impedance Calculator

Table of Contents

\( \) \( \) \( \)

A calculator to calculate the equivalent impedance of a resistor and an inductor in series. The impedance is given as a complex number in standard form and polar forms.

Formulae for Series R L Circuit Impedance Used in Calculator and their Units

series R L circuit

Let \( f \) be the frequency, in Hertz, of the source voltage supplying the circuit.
and define the following parameters used in the calculations
\( \omega = 2 \pi f \) , angular frequency in rad/s
\( X_L = \omega L \) , the inductive reactance in ohms \( (\Omega) \)
The impedance of the inductor \( L \) is given by
\( Z_L = j \omega L \)
Let \( Z \) be the equivalent impedance to the series RL circuit shown above and write it in complex form as follows
\[ Z = R + Z_L = R + j\omega L \]
The formulae for the modulus \( |Z| \) and argument (or phase) \( \theta \) of \( Z \) are given by

Modulus: \( |Z| = \sqrt{ R^2 + \omega^2 L^2 } \) in ohms \( (\Omega) \)

Argument (Phase): \( \theta = \arctan ( \dfrac{\omega L }{R} ) \) in radians or degrees

Use of the calculator

Enter the resistance, the capacitance and the frequency as positive real numbers with the given units then press "calculate".

Results of Calculations

    
    
    
    
    

More References and links

AC Circuits Calculators and Solvers
Complex Numbers - Basic Operations
Complex Numbers in Exponential Form
Complex Numbers in Polar Form
Convert a Complex Number to Polar and Exponential Forms Calculator
Engineering Mathematics with Examples and Solutions
Sours: http://www.mathforengineers.com/AC-circuits-calculators/series-RL-circuit-Impedance.html
RL Circuit Analysis (1 of 8) Voltage and Current
Scheme

This series RL circuit impedance calculator determines the impedance and the phase difference angle of an inductor and a resistor connected in series for a given frequency of a sinusoidal signal. The angular frequency is also determined.

Example:Calculate the impedance of a 500 mH inductor and a 0.2 Ω resistor at a frequency of 25 kHz.

Input

Resistance, R

Inductance, L

Frequency, f

Share

Output

Angular Frequencyω= rad/s

Inductive reactanceXL= Ω

Total RL Impedance|ZRL|= Ω

Phase differenceφ = ° = rad

Enter the resistance, inductance, and frequency values, select the units and click or tap the Calculate button. Try to enter zero or infinitely large values to see how this circuit behaves. Infinite frequency is not supported. To enter the Infinity value, just type inf in the input box.

The following formulas are used for the calculation:

Formula

Formula

Formula

Formula

Formula

where

ZRL is the RL circuit impedance in ohms (Ω),

ω = 2πf is the angular frequency in rad/s,

f is the frequency in hertz (Hz),

R is the resistance in ohms (Ω),

L is the inductance in henries (H), and

φ is the phase difference between the total voltage VTand the total current IT in degrees (°) and radians, and

j is the imaginary unit.

Picture

A graph of the series RL circuit impedance ZRL against frequency f for a given inductance and resistance

To calculate, enter the inductance, the resistance, and the frequency, select the units of measurements and the result for RL impedance will be shown in ohms and for the phase difference in degrees. The inductive reactance in ohms will also be calculated.

The phasor diagram for a series RL circuit shows that the total current wave lags behind the total voltage wave. The lag is less than 90° and more than 0°. At 90° a jumper is installed instead of a resistor (the circuit is purely inductive) and at 0° a jumper is installed instead of the inductor (the circuit is purely resistive)

The phasor diagram for a series RL circuit shows that the total current wave lags behind the total voltage wave. The lag is less than 90° and more than 0°. At 90° a jumper is installed instead of a resistor (the circuit is purely inductive) and at 0° a jumper is installed instead of the inductor (the circuit is purely resistive)

A simple series RL or resistor-inductor circuit is composed of a resistor and an inductor connected in series and driven by a voltage source. The current in both inductor and resistor is the same because they are connected in series. The voltages across the resistor VR and the inductor VL are shown in the diagram at the right angle to each other. Their sum is always greater than the total voltage VT.

If you look at the equation for calculating the impedance (above), you will notice that it looks like the equation for calculating the hypotenuse of a right triangle. This is because the impedance of an RL circuit in graphical form looks like in this picture where the resistance R is on the horizontal axes and the reactance XL is on the vertical axis. The hypotenuse is the impedance of the circuit and the phase angle is the angle between the horizontal axis and the impedance vector.

The phase angle can be between 0° for a purely resistive circuit and 90° for a purely inductive circuit. From the reactance triangle

Formula

or using the arctangent (inverse tangent) function

Formula

Picture

In a series RL circuit with the sine voltage source, the current wave lags behind the voltage wave with the lag less than 90° (zero resistance) and more than 0° (zero inductance). In other words, voltage leads current by φ in phase; 0° ≤ φ ≤ 90°. If the voltage V is V = Vmsin(2πft), then the current I is I = Imsin(2πft + φ) where Vm and Im are voltage and current amplitudes, f is the frequency (constant), φ is the phase angle (constant), and t is the time (variable)

In a series RL circuit, the same current I flows through both the inductor and the resistor. The inductor’s voltage VLleads the common current by 90° and the resistor voltage is in phase with the common current. From Kirchhoff's voltage law, the sum of the voltage drops must equal the total voltage VT. Resistor and inductor voltages VR and VL are 90° out of phase with each other, therefore they must be added as phasor quantities and the total voltage VT is determined as

Formula

Note that the total voltage is always less than the sum of the voltages across the resistor and the inductor — exactly like in any right-angle triangle where the length of the hypotenuse is shorter than the sum of the two legs of the triangle.

Note also that it is impossible to measure impedance directly using an ordinary multimeter — you have to use an impedance meter for this purpose. An example of use is a measurement of the impedance of several speakers with transformers, voice coils, and crossovers. Unlike a multimeter, which applies direct voltage to the circuit being measured, the impedance meter applies the AC test signal to the circuit being tested.

Failure Modes

What if something goes wrong in this circuit? Click or tap a corresponding link to view the calculator in various failure modes:

Special Modes

Click or tap a corresponding link to view the calculator in various special modes:

Various direct current modes

Short circuit

Open circuit

Purely inductive circuit

Inductive circuit

Notes

  • In our explanations of the behavior of this circuit, zero frequency means direct current. If f = 0, we assume that the circuit is connected to an ideal DC voltage source.
  • At zero frequency, we consider the reactance of an ideal inductor to be infinitely large if its inductance is infinitely large. If the inductor has a finite inductance, its reactance at zero frequency is zero and for a DC voltage source, it represents a short circuit.

This article was written by Anatoly Zolotkov

CalculatorsElectrical, RF and Electronics Calculators

Sours: https://www.translatorscafe.com/unit-converter/en-US/calculator/series-rl-impedance/

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