Such as, Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. In case of series RL circuit, resistor and inductor are connected in series, so current flowing in both the elements are same i.e I R = I L = I. The above diagram shows the impedance triangle.
For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. i.e VR is in phase with I. Step- III. The impedance of series RL circuit opposes the flow of alternating current. We are a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for us to earn fees by linking to Amazon.com and affiliated sites. A circuit that contains a pure resistance R ohms connected in series with a coil having a pure inductance of L (Henry) is known as RL Series Circuit. When an AC supply voltage V is applied, the current, I flows in the circuit. If the time of the measurement were much larger than the time constant, we would not see the decay or growth of the voltage across the inductor or resistor. This then forms the basis of an RL charging circuit were 5 τ can also be thought of as “ 5*(L/R) ” or the transient time of the circuit. Power triangle for series RL circuit is shown below, RL Series Circuit A circuit that contains a pure resistance R ohms connected in series with a coil having a pure inductance of L (Henry) is known as RL Series Circuit.When an AC supply voltage V is applied, the current, I flows in the circuit. We can also use that same relationship as a substitution for the energy in an inductor formula to find how the energy decreases at different time intervals. (b) How long does it take before the energy stored in the inductor is reduced to \(1.0 \%\) of its maximum value? (b) What are the current in the circuit and the magnitude of the induced emf across the inductor at \(t = 0,\) at \(t = 2.0 \tau_L\), and as \(t \rightarrow \infty\)? Substituting the values we get, Before drawing the phasor diagram of series RL circuit, one should know the relationship between voltage and current in case of resistor and inductor. The time constant \(\tau_L\) tells us how rapidly the current increases to its final value. Inductive reactance also increases as it is directly proportional to frequency. When \(S_1\) is opened and \(S_2\) is closed, the circuit becomes a single-loop circuit with only a resistor and an inductor (Figure \(\PageIndex{1c}\)). Nicely explained I have solved my doubts by reading this. Calculate the voltages across resistor R and inductor L by using Ohm’s Law. The term L/R in the equation is called the Time Constant, (τ) of the RL series circuit, and it is defined as time taken by the current to reach its maximum steady state value and the term V/R represents the final steady state value of current in the circuit. It is measured in ohms (Ω).
Step 3. As a result, I(t) starts at zero and increases asymptotically to its final value. [ "article:topic", "authorname:openstax", "inductive time constant", "RL circuits", "license:ccby", "showtoc:no" ], Creative Commons Attribution License (by 4.0), Analyze circuits that have an inductor and resistor in series, Describe how current and voltage exponentially grow or decay based on the initial conditions, The inductive time constant is \[\tau_L = \dfrac{L}{R} = \dfrac{4.0 \, H}{4.0 \, \Omega} = 1.0 \, s.\], The current in the circuit of Figure \(\PageIndex{1b}\) increases according to Equation \ref{eq3}: \[I(t) = \dfrac{\epsilon}{R}(1 - e^{-t/\tau_L}).\] At \(t = 0\), \[(1 - e^{-t/\tau_L}) = (1 - 1) = 0; \, so \, I(0) = 0.\] At \(t = 2.0 \tau_L\) and \(t \rightarrow \infty\), we have, respectively, \[I(2.0\tau_L) = \dfrac{\epsilon}{R}(1 - e^{-2.0}) = (0.50 \, A)(0.86) = 0.43 \, A,\] and \[I(\infty) = \dfrac{\epsilon}{R} = 0.50 \, A.\] From Equation \ref{14.32}, the magnitude of the induced emf decays as \[|V_L(t)| = \epsilon e^{-t/\tau_L}.\] At \(t = 0\), \(t = 2.0 \tau_L\), and as \(t \rightarrow \infty\), we obtain \[|V_L(0)| = \epsilon = \, V,\] \[|V_L(2.0 \tau_L)| = (2.0 \, V) e^{-2.0} = 0.27 \, V\] and \[|V_L(\infty)| = 0.\], With the switches reversed, the current decreases according to \[I(t) = \dfrac{\epsilon}{R}e^{-t/\tau_L} = I(0)e^{-t/\tau_L}.\] At a time, The energy stored in the inductor is given by \[U_L(t) = \dfrac{1}{2}L[I(t)]^2 = \dfrac{1}{2}L\left(\dfrac{\epsilon}{R} e^{-t/\tau_L}\right)^2 = \dfrac{L\epsilon^2}{2R^2}e^{-2t/\tau_L}.\] If the energy drops to \(1.0\%\) of its initial value at a time. The current lags the voltage and thus they are not in phase with each other. Figure \(\PageIndex{1a}\) shows an RL circuit consisting of a resistor, an inductor, a constant source of emf, and switches \(S_1\) and \(S_2\). Let us say at time ‘t’ we close the switch and the current ‘i’ starts flowing in the circuit but it does not attains its maximum value rapidly due to the presence of inductor in the circuit as we know inductor has a property to oppose the change in the current flowing through it. Similarly, the solution to Equation \ref{eq1} can be found by making substitutions in the equations relating the capacitor to the inductor. It starts at zero, and as \(t \rightarrow \infty\), I(t) approaches \(\epsilon/R\) asymptotically. An RL circuit (also known as an RL filter or RL network) is defined as an electrical circuit consisting of the passive circuit elements of a resistor (R) and an inductor (L) connected together, driven by a voltage source or current source. It is given by the equation: If the alternating voltage applied across the circuit is given by the equation: Then the instantaneous power is given by the equation: Putting the value of v and i from the equation (1) and (2) in the equation (3) we will get. Integrating both sides, we get, Otherwise, we start with a lower initial current, which will decay by the same relationship. Once \(S_1\) is closed and \(S_2\) is open, the source of emf produces a current in the circuit. The smaller the inductive time constant \(\tau_L = L/R\), the more rapidly the current approaches \(\epsilon/R\). Thus, as the current approaches the maximum current \(\epsilon/R\), the stored energy in the inductor increases from zero and asymptotically approaches a maximum of \(L(\epsilon/R)^2 /2\). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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