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14.4: RL Circuits

A circuit with resistance and self-inductance is known as an RL circuit. Figure \(\PageIndex{1a}\) shows an RL circuit consisting of a resistor, an inductor, a constant source of emf, and switches \(S_1\) and \(S_2\). When \(S_1\) is closed, the circuit is equivalent to a single-loop circuit consisting of a resistor and an inductor connected across a source of emf (Figure \(\PageIndex{1b}\)). 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 \(\PageIndex{1c}\)).

Figure a shows a resistor R and an inductor L connected in series with two switches which are parallel to each other. Both switches are currently open. Closing switch S1 would connect R and L in series with a battery, whose positive terminal is towards L. Closing switch S2 would form a closed loop of R and L, without the battery. Figure b shows a closed circuit with R, L and the battery in series. The side of L towards the battery, is at positive potential. Current flows from the positive end of L, through it, to the negative end. Figure c shows R and L connected in series. The potential across L is reversed, but the current flows in the same direction as in figure b

Figure \(\PageIndex{1}\): (a) An RL circuit with switches \(S_1\) and \(S_2\). (b) The equivalent circuit with \(S_1\) closed and \(S_2\) open. (c) The equivalent circuit after \(S_1\) is opened and \(S_2\) is closed.

We first consider the RL circuit of Figure \(\PageIndex{1b}\). Once \(S_1\) is closed and \(S_2\) is open, the source of emf produces a current in the circuit. If there were no self-inductance in the circuit, the current would rise immediately to a steady value of \(\epsilon/R\). However, from Faraday’s law, the increasing current produces an emf \(V_L = -L(dI/dt)\) across the inductor. In accordance with Lenz’s law, the induced emf counteracts the increase in the current and is directed as shown in the figure. As a result, I(t) starts at zero and increases asymptotically to its final value.

Applying Kirchhoff’s loop rule to this circuit, we obtain

\[\epsilon - L\dfrac{dI}{dt} - IR = 0, \label{eq1}\]

which is a first-order differential equation for \(I(t)\). Notice its similarity to the equation for a capacitor and resistor in series (see RC Circuits). Similarly, the solution to Equation \ref{eq1} can be found by making substitutions in the equations relating the capacitor to the inductor. This gives

\[ \begin{align} I(t) &= \dfrac{\epsilon}{R}(1 - e^{-Rt/L}) \\[5pt] &= \dfrac{\epsilon}{R}(1 - e^{-t/\tau_L}), \end{align} \label{eq3}\]

where

\[\tau_L = L/R \label{eq5}\]

is the inductive time constant of the circuit.

The current \(I(t)\) is plotted in Figure \(\PageIndex{2a}\). It starts at zero, and as \(t \rightarrow \infty\), I(t) approaches \(\epsilon/R\) asymptotically. The induced emf \(V_L(t)\) is directly proportional to dI/dt, or the slope of the curve. Hence, while at its greatest immediately after the switches are thrown, the induced emf decreases to zero with time as the current approaches its final value of \(\epsilon/R\). The circuit then becomes equivalent to a resistor connected across a source of emf.

 Figure a shows the graph of electric current I versus time t. Current increases with time in a curve which flattens out at epsilon I R. At t equal to tau subscript L, the value of I is 0.63 epsilon I R. Figure b shows the graph of magnitude of induced voltage, mod V subscript L, versus time t. Mod V subscript L starts at value epsilon and decreases with time till the curve reaches zero. At t equal to tau subscript L, the value of I is 0.37 epsilon

Figure \(\PageIndex{2}\): Time variation of (a) the electric current and (b) the magnitude of the induced voltage across the coil in the circuit of Figure \(\PageIndex{1b}\).

The energy stored in the magnetic field of an inductor is

\[U_L = \dfrac{1}{2}LI^2.\]

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 \(L(\epsilon/R)^2 /2\).

The time constant \(\tau_L\) tells us how rapidly the current increases to its final value. At \(t = \tau_L\), the current in the circuit is, from Equation \ref{eq3},

\[I(\tau_L) = \dfrac{\epsilon}{R}(1 - e^{-1}) = 0.63 \dfrac{\epsilon}{R},\]

which is \(63\%\) of the final \(\epsilon/R\). The smaller the inductive time constant \(\tau_L = L/R\), the more rapidly the current approaches \(\epsilon/R\).

We can find the time dependence of the induced voltage across the inductor in this circuit by using \(V_L(t) = -L(dI/dt)\) and Equation \ref{eq3}:

\[V_L(t) = -L\dfrac{dI}{dt} = -\epsilon e^{-t/\tau_L}.\]

The magnitude of this function is plotted in Figure \(\PageIndex{2b}\). The greatest value of \(L(dI/dt)\) is \(\epsilon\); it occurs when dI/dt is greatest, which is immediately after \(S_1\) is closed and \(S_2\) is opened. In the approach to steady state, dI/dt decreases to zero. As a result, the voltage across the inductor also vanishes as \(t \rightarrow \infty\).

The time constant \(\tau_L\) also tells us how quickly the induced voltage decays. At \(t = \tau_L\) the magnitude of the induced voltage is

\[|V_L(\tau_L)| = \epsilon e^{-1} = 0.37 \epsilon = 0.37 V(0).\]

The voltage across the inductor therefore drops to about \(37\%\) of its initial value after one time constant. The shorter the time constant \(\tau_L\), the more rapidly the voltage decreases.

After enough time has elapsed so that the current has essentially reached its final value, the positions of the switches in Figure \(\PageIndex{1a}\) are reversed, giving us the circuit in part (c). At \(t = 0\), the current in the circuit is \(I(0) = \epsilon/R\). With Kirchhoff’s loop rule, we obtain

\[IR + L\dfrac{dI}{dt} = 0.\]

The solution to this equation is similar to the solution of the equation for a discharging capacitor, with similar substitutions. The current at time t is then

\[I(t) = \dfrac{\epsilon}{R}e^{-t/\tau_L}.\]

The current starts at \(I(0) = \epsilon/R\) and decreases with time as the energy stored in the inductor is depleted (Figure \(\PageIndex{3}\)).

The time dependence of the voltage across the inductor can be determined from \(V_L = -L(dI/dt)\):

\[V_L(0) = \epsilon e^{-t/\tau_L}.\]

This voltage is initially \(V_L(0) = \epsilon\), and it decays to zero like the current. The energy stored in the magnetic field of the inductor, \(LI^2/2\), also decreases exponentially with time, as it is dissipated by Joule heating in the resistance of the circuit.

The graph of I versus t. The value of I at t equal to 0 is epsilon I R. I decreases with time till the curve reaches 0. At t equal to tau subscript L, the value of I is 0.37 epsilon I R.

Figure \(\PageIndex{3}\): Time variation of electric current in the RL circuit of Figure(c). The induced voltage across the coil also decays exponentially.

Example \(\PageIndex{1}\): An RL Circuit with a Source of emf

In the circuit of Figure \(\PageIndex{1a}\), let \(\epsilon = 2.0 \space V, \space R = 4.0 \space \Omega\), and \(L = 4.0 \space H\). With \(S_1\) closed and \(S_2\) open (Figure \(\PageIndex{1b}\)), (a) what is the time constant of the circuit? (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 \(t \rightarrow \infty\)?

Strategy

The time constant for an inductor and resistor in a series circuit is calculated using Equation. The current through and voltage across the inductor are calculated by the scenarios detailed from Equation and Equation.

Solution

  1. The inductive time constant is \[\tau_L = \dfrac{L}{R} = \dfrac{4.0 \space H}{4.0 \space \Omega} = 1.0 \space s.\]
  2. The current in the circuit of Figure(b) increases according to Equation: \[I(t) = \dfrac{\epsilon}{R}(1 - e^{-t/\tau_L}).\] At \(t = 0\), \[(1 - e^{-t/\tau_L}) = (1 - 1) = 0; \space so \space I(0) = 0.\] At \(t = 2.0 \tau_L\) and \(t \rightarrow \infty\), we have, respectively, \[I(2.0\tau_L) = \dfrac{\epsilon}{R}(1 - e^{-2.0}) = (0.50 \space A)(0.86) = 0.43 \space A,\] and \[I(\infty) = \dfrac{\epsilon}{R} = 0.50 \space A.\] From Equation, 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 \[|V_L(0)| = \epsilon = \space V,\] \[|V_L(2.0 \tau_L)| = (2.0 \space V) e^{-2.0} = 0.27 \space V\] and \[|V_L(\infty)| = 0.\]

Significance

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. The circuit would quickly reach the asymptotic values for both of these (Figure \(\PageIndex{4}\)).

 Figures a, b and c show the oscilloscope traces of voltage versus time of the voltage across source, the voltage across the inductor and the voltage across the resistor respectively. Figure a is a square wave varying from minus 12 volts to plus 12 volts, with a period from minus 10 ms to minus 0.001 ms. Figure b shows a square wave varying from minus 6 volts to plus 6 volts with a spike of 16 volts at the beginning of every crest and a spike of minus 16 volts at the beginning of every trough. The period is the same as that in figure a. Figure c shows a square wave varying from minus 0.3 to plus 0.3 volts, with spikes going out of the trace area in the positive direction at the beginnings of every crest and trough. The period of the wave is from minus 9.985 to plus 0.015 ms.

Figure \(\PageIndex{4}\): A generator in an RL circuit produces a square-pulse output in which the voltage oscillates between zero and some set value. These oscilloscope traces show (a) the voltage across the source; (b) the voltage across the inductor; (c) the voltage across the resistor.

Example \(\PageIndex{1}\): An RL Circuit without a Source of emf

After the current in the RL circuit of Example \(\PageIndex{1}\) has reached its final value, the positions of the switches are reversed so that the circuit becomes the one shown in Figure(c). (a) How long does it take the current to drop to half its initial value? (b) How long does it take before the energy stored in the inductor is reduced to \(1.0 \%\) of its maximum value?

Strategy

The current in the inductor will now decrease as the resistor dissipates this energy. Therefore, the current falls as an exponential decay. 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.

Solution

  1. 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 t when the current is one-half its initial value, we have \[I(t) = 0.50 I(0)\space so \space e^{-t/\tau_L} = 0.50,\] and \[t = -[ln(0.50)]\tau_L = 0.69 (1.0 \space s) = 0.69 \space s,\] where we have used the inductive time constant found in Example.
  2. 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 t, we have \[U_L(t) = (0.010)U_L(0) \space or \space \dfrac{L\epsilon^2}{2R^2} e^{-2t/\tau_L} = (0.010)\dfrac{L\epsilon^2}{2R^2}.\] Upon canceling terms and taking the natural logarithm of both sides, we obtain \[-\dfrac{2t}{\tau_L} = ln(0.010),\] so \[t = -\dfrac{1}{2} \tau_L ln(0.010).\] Since \(\tau_L = 1.0 \space s\), the time it takes for the energy stored in the inductor to decrease to \(1.0\%\) of its initial value is \[t = -\dfrac{1}{2}(1.0 \space s)ln(0.010) = 2.3 \space s.\]

Significance

This calculation only works if the circuit is at maximum current in situation (b) prior to this new situation. Otherwise, we start with a lower initial current, which will decay by the same relationship.

Note

Check Your Understanding Verify that RC and L/R have the dimensions of time.

Note

Check Your Understanding (a) If the current in the circuit of in Figure(b) increases to \(90\%\) of its final value after 5.0 s, what is the inductive time constant? (b) If \(R = 20 \space \Omega\), what is the value of the self-inductance? (c) If the \(R = 20 \space \Omega\) resistor is replaced with a \(R = 100 \space \Omega\) resistor, what is the time taken for the current to reach \(90\%\) of its final value?

[Hide Solution]

a. 2.2 s; b. 43 H; c. 1.0 s

Note

Check Your Understanding For the circuit of in Figure(b), show that when steady state is reached, the difference in the total energies produced by the battery and dissipated in the resistor is equal to the energy stored in the magnetic field of the coil.

Contributors

Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).