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All coils, inductors, chokes and transformers create a magnetic field around themselves consist of an Inductance in series with a Resistance forming an LR Series Circuit
Inductive coils and solenoids are not purely inductive devices, but instead consist of an inductance and resistance connected together to form a basic LR series circuit.
In the first tutorial in this section about Inductors, we looked briefly at the time constant of an inductor stating that the current flowing through an inductor could not change instantaneously, but would increase at a constant rate determined by the self-induced back-emf within the inductive coil.
In other words, an inductor in an electrical circuit opposes the flow of current, ( i ) through it. While this is perfectly correct, we made the assumption in the tutorial that it was an ideal inductor which had no resistance or capacitance associated with its coil windings.
However, in the real world “ALL” coils whether they are chokes, solenoids, relays or any wound component will always have a certain amount of resistance no matter how small. This is because the actual coils turns of wire being used to make it uses copper wire which has a resistive value.
Then for real world purposes we can consider our simple coil as being an “Inductance”, L in series with a “Resistance”, R. In other words forming an LR Series Circuit.
A LR Series Circuit consists basically of an inductor of inductance, L connected in series with a resistor of resistance, R. The resistance “R” is the DC resistive value of the wire turns or loops that goes into making up the inductors coil. Consider the LR series circuit below.
The above LR series circuit is connected across a constant voltage source, (the battery) and a switch. Assume that the switch, S is open until it is closed at a time t = 0, and then remains permanently closed producing a “step response” type voltage input. The current, i begins to flow through the circuit but does not rise rapidly to its maximum value of Imax as determined by the ratio of V / R (Ohms Law).
This limiting factor is due to the presence of the self induced emf within the inductor as a result of the growth of magnetic flux, (Lenz’s Law). After a time the voltage source neutralizes the effect of the self induced emf, the current flow becomes constant and the induced current and field are reduced to zero.
We can use Kirchhoff’s Voltage Law, (KVL) to define the individual voltage drops that exist around the circuit and then hopefully use it to give us an expression for the flow of current.
Kirchhoff’s voltage law (KVL) gives us:
The voltage drop across the resistor, R is I*R (Ohms Law).
The voltage drop across the inductor, L is by now our familiar expression L(di/dt)
Then the final expression for the individual voltage drops around the LR series circuit can be given as:
We can see that the voltage drop across the resistor depends upon the current, i, while the voltage drop across the inductor depends upon the rate of change of the current, di/dt. When the current is equal to zero, ( i = 0 ) at time t = 0 the above expression, which is also a first order differential equation, can be rewritten to give the value of the current at any instant of time as:
The Time Constant, ( τ ) of the LR series circuit is given as L/R and in which V/R represents the final steady state current value after five time constant values. Once the current reaches this maximum steady state value at 5τ, the inductance of the coil has reduced to zero acting more like a short circuit and effectively removing it from the circuit.
Therefore the current flowing through the coil is limited only by the resistive element in Ohms of the coils windings. A graphical representation of the current growth representing the voltage/time characteristics of the circuit can be presented as.
Since the voltage drop across the resistor, VR is equal to I*R (Ohms Law), it will have the same exponential growth and shape as the current. However, the voltage drop across the inductor, VL will have a value equal to: Ve(-Rt/L). Then the voltage across the inductor, VL will have an initial value equal to the battery voltage at time t = 0 or when the switch is first closed and then decays exponentially to zero as represented in the above curves.
The time required for the current flowing in the LR series circuit to reach its maximum steady state value is equivalent to about 5 time constants or 5τ. This time constant τ, is measured by τ = L/R, in seconds, where R is the value of the resistor in ohms and L is the value of the inductor in Henries. This then forms the basis of an RL charging circuit were 5τ can also be thought of as “5*(L/R)” or the transient time of the circuit.
The transient time of any inductive circuit is determined by the relationship between the inductance and the resistance. For example, for a fixed value resistance the larger the inductance the slower will be the transient time and therefore a longer time constant for the LR series circuit. Likewise, for a fixed value inductance the smaller the resistance value the longer the transient time.
However, for a fixed value inductance, by increasing the resistance value the transient time and therefore the time constant of the circuit becomes shorter. This is because as the resistance increases the circuit becomes more and more resistive as the value of the inductance becomes negligible compared to the resistance. If the value of the resistance is increased sufficiently large compared to the inductance the transient time would effectively be reduced to almost zero.
A coil which has an inductance of 40mH and a resistance of 2Ω is connected together to form a LR series circuit. If they are connected to a 20V DC supply.
a). What will be the final steady state value of the current.
b) What will be the time constant of the RL series circuit.
c) What will be the transient time of the RL series circuit.
d) What will be the value of the induced emf after 10ms.
e) What will be the value of the circuit current one time constant after the switch is closed.
The Time Constant, τ of the circuit was calculated in question b) as being 20ms. Then the circuit current at this time is given as:
You may have noticed that the answer for question (e) which gives a value of 6.32 Amps at one time constant, is equal to 63.2% of the final steady state current value of 10 Amps we calculated in question (a). This value of 63.2% or 0.632 x IMAX also corresponds with the transient curves shown above.
Then from above, the instantaneous rate at which the voltage source delivers power to the circuit is given as:
The instantaneous rate at which power is dissipated by the resistor in the form of heat is given as:
The rate at which energy is stored in the inductor in the form of magnetic potential energy is given as:
Then we can find the total power in a RL series circuit by multiplying by i and is therefore:
Where the first I2R term represents the power dissipated by the resistor in heat, and the second term represents the power absorbed by the inductor, its magnetic energy.
I think you need to look at this in depth
still good
Nice explanation
I like this moment
I wish all the best
At t=0, switch is suddenly closed. The electric current is zero. The voltage across inductor should be zero.
Inductor voltage will increase to maximum which is less than applied voltage because the voltage across the resistor. Inductor voltage then drop slowly to zero.
Therefore, your exponential function is incorrect.
I think you are backwards. The voltage across the inductor will be equal to the applied voltage at T-0. Because at that instant, there is no current. And you cannot drop voltage without current. V=IxR. As the current slowly starts to flow through the inductor, then the voltage will drop to zero over 5T and current will rise to max over 5T, and the inductor’s opposition to current will be reduced to whatever it’s actual resistance is. The voltage across the inductor will be equal to the applied voltage because the resistor is not a separate component, it is a representation of the coils resistance.
Inductance, like ordinary mass, possesses inertia. It is not possible to instantaneously alter the current in an inductor, unless infinite voltages are involved. With good reason, Maxwell called the energy of an inductor Electrokinetic Energy, and the product of the inductance by its current, Electrokinetic Momentum. Hence, when the switch is closed, there is no current, no voltage across the resistor, and the whole of the input voltage is impressed upon the inductor. As the current grows, so too does the voltage across the resistor, gradually decreasing the rate of increase of current. Please remember this for the rest of your electronic life!
When subjected to a step input change at time t = 0, the inductor current IL(t) exponentially increases from 0 to V/R at a time constant given by: τ = L/R seconds. At the same time the voltage VL(t) exponentially decreases from V to 0 at the same time constant rate. When elapsed time is greater than 5 time constants, 5T the inductor will act as a short circuit, thus IL is at its maximum value limited only by R, and VL is zero as correctly shown in the tutorial.
Good work
Thank you l have been helped a lot
So I derived the equation in yellow “Expression for the Current in an LR Series Circuit” myself for practice and I was confused by (l + e^((R/L)t)). That’s a lower case “L” right, not a 1? It should be an upper case L so as not to be confusing right?
If it’s a 1 I am REALLY confused.
Yes, an L(or so I hope).
When a DC step voltage is applied to a coil consisting of an inductance and resistance in series, the current flowing through this RL series cicruit does not rise in value instantaneously, but the current increases as an exponential curve because the magnetic field being created by the inductance of the coil opposes any change in current through it.
The exponential curve of the current from zero to its steady state value depends greatly on the RL time constant (τ = L/R), thus the formula given for the exponential rise in current through the coil at any instant in time is given as:
i(t) = (V/R)*( 1 – e-Rt/L) as shown in the tutorial.
This information was helpful
This was of great importance… I’m really impressed ❤️😋😋
À simple R-L circuit is excited with a constant voltage source the speed of the response depends on what?
how to calculate steady state error for RL circuit
It’s a great job the experiment is correct
How do i calculate the winding resistance when given only R, L and frequency.
Please read the tutorial about Inductive Reactance
Curves for transients in L-R circuits in both induced voltage and resistor voltage
How can I find out the power dissipated by the resistor in equilibrium?
P = V*I = I2R = V2/R
I have been assisted greatly in forming derivatives of RL circuit.thanks to the administrators.
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Please explian what happens when the volatage drop across both resistance and indactance are same
If there is the same value of voltage drop across each component due to the series current, then R = XL
when you wind a wire on an iron rod and connect both ends to a battery positive and negative terminal to magnetize the rod , why doesn’t the battery shorted out , even though it is winded still the wire is directly connected from positive to negative ?
Thanks ;
Jacob
The wire has resistance