# RL Circuits Problems with Answers

In this tutorial, RL circuits and related formulas are discussed, and then some problems with answers are presented that are useful for high schools.

## RL Series Circuit Fact Sheet:

Any circuit including a resistor, an inductance, and an emf that are in series or parallel is called an RL circuit.

To solve RL circuit problems, you should apply Kirchhoff's voltage rule to find the necessary equations.

An inductor with self-inductance $L$ is used in a single-loop circuit to stop the current from reaching its maximum value instantaneously.

## RL Circuits Formulas

Case I: If we have a single-loop circuit in which components $R$ and $L$ are connected to a battery with emf $\mathcal{E}$, then the current in this RL circuit is obtained by formula below $I=\frac{\mathcal{E}}{R}\left(1-e^{(-Rt/L)}\right)=I_{max}\left(1-e^{-t/\tau}\right)$ Where:
$\mathcal{E}$ is emf of the battery in Volts
$L$ is the self-inductance of the inductor in Henries
$R$ is the resistance of the resistor in Ohms
$e$ is the base of Natural Logarithm, $e=2.71828$
$\tau$ that is called the time constant has units of time in seconds and is the time it takes the current in the circuits to reach $63.2\%$ of its maximum (final) value $I_{max}$. It defines as $\tau=\frac{L}{R}$.

Note: for an RC circuit, the time constant is defined as $\tau=RC$.

Case II: If the source of energy, the battery, is removed from the circuit by opening a switch, then the current does not drop to zero through the resistor instantaneously and takes some time. In this case, the current at any time is obtained using the below formula $I=I_0 e^{-t/\tau}$ where $I_0$ is the initial current before the switch is closed.

## RL Circuit Solved Problems

Problem (1): A solenoid with an inductance of 25 mH and resistance of ${8\,\rm \Omega}$ are connected to the terminals of a 6-V battery in series. There is also a switch in the circuit.
(a) Immediately after the switch is closed, find the potential drop across the resistor.
(b) Find the final current in the circuit.
(c) Find the time constant of the circuit.
(d) Find the current after one time constant has elapsed.

Solution: Recall that the current in an RL circuit at any instant of time is obtained by the formula $I(t)=\frac{\mathcal{E}}{R}(1-e^{-Rt/L})$
(a) The potential drop across the resistor in an RL circuit is given by $\Delta V=IR$, so by setting $t=0$ in the above equation for current, we can find the current just after closing the switch as below $I(t=0)=\frac{\mathcal{E}}{R}(1-e^{-R(0)/L})=0$ Thus, the voltage drop across the resistor is $\Delta V_R=T(t=0)R=0$ immediately after the switch is closed.

(b) In an RL circuit, the final or maximum current is found by the formula $I_f=\frac{\mathcal{E}}{R}$. The final current in this single-loop circuit is $I_f=\frac{6}{8}=0.75\quad {\rm A}$
(c) By definition, the time constant in an RL circuit is the time that takes the current to reach $63.2\%$ of its final value. Therefore, $\tau=\frac LR=\frac {25\times 10^{-3}}{8}=3\,{\rm ms}$
(d) the current after on time constant, $I(t=\tau)$ is calculated as below \begin{align*} I(t)&=\frac{\mathcal{E}}{R}(1-e^{-t/\tau})\\ \\&=\frac{6}{8}(1-e^{-\tau/\tau})\\ \\&=0.75(1-e^{-1})\\ \\&=0.474\quad {\rm A}\end{align*}

Be sure to check the following article before taking your AP Physics exam:

AP Physics Circuits Practice Problems with Solutions

Problem (2): A $5\,{\rm mH}$-inductor, a $15\,{\rm \Omega}$-resistor are connected across a $12\,{\rm V}$- battery with negligible internal resistance in series.
(a) What is the final current in the circuit?
(b) What is the time constant?
(c) What is the current after two time constants have elapsed?
(d) What is the potential drop across the resistance after two time constants?
(e) What is the potential drop across the inductor after two time constants?

Solution: Again, the master formula in the RL circuits when connected to a source of energy (battery) is $I(t)=\frac{\mathcal{E}}{R}(1-e^{-Rt/L})$
(a) final current is the current when there is no variation in the current i.e. $\frac{dI}{dt}=0$ which is obtained as $I_f=\frac{\mathcal{E}}{R}$. Thus, $I_f=\frac{12}{15}=0.8\quad {\rm A}$
(b) the time when the current reaches $63.2\%$ of its final value is the time constant of the RL circuit i.e. $I(t=\tau)=0.632I_f$. Thus, we have $\tau=\frac{L}{R}=\frac{5\times 10^{-3}}{15}=333\quad {\rm \mu s}$ where $\mu$ denoted microseconds, $1\mu s=10^{-6}\,{\rm s}$.

(c) Setting $t=2\tau$ in the current equation above, we get  \begin{align*} I(t=2\tau)&=I_f(1-e^{-t/\tau})\\ \\&=0.8(1-e^{-2\tau/\tau})\\ \\&=0.8(1-e^{-2})\\ \\&=0.691\quad {\rm A}\end{align*} The meaning of this expression is that after $666\,{\rm \mu s}$, the current reaches $86.4\%$ of its final value which is $0.691\,{\rm A}$.

(d) Potential difference across the resistor is $\Delta V_R=I(t)R$. Thus, after two time constant, we have \begin{align*}\Delta V_R&=I(t=2\tau)R\\&=0.691\times 15 \\&=10.365\quad{\rm V}\end{align*}
(e) Applying Kirchhoff's voltage rule around the circuit, we obtain $\mathcal{E}-\Delta V_R-\Delta V_L=0$ where $\Delta V_L$ is the voltage across the inductor at any instant of time. In the previous part, we calculated the $\Delta V_R$ after two time constants, so we have \begin{gather*} \mathcal{E}-\Delta V_R-\Delta V_L=0 \\ 12-10.36-\Delta V_L=0 \\ \Rightarrow \Delta V_L=1.635\quad{\rm V}\end{gather*} Note that $\mathcal{E}$ is the emf of the battery in the circuit.

Problem (3): Consider a series circuit including a $4\,{\rm \Omega}$, a $16\,{\rm mH}$-inductor, and a power supply of 9 volts.
(a) Find the time constant of the RL series circuit?
(b) Find the current in the circuit  $100\,{\rm \mu s}$ after the switch is closed?
(c) How many time constants does it take for the current to reach $99.9\%$ of its final value?

Solution: the given data is $R=4\,{\rm \Omega}$, $L=16\,{\rm mH}$, $\mathcal{E}=9\,{\rm V}$.
(a) Time constant is the ratio of the inductance to the resistance of an RL circuit. Thus, $\tau=\frac{L}{R}=\frac{16\times 10^{-3}}{4}=4\,{\rm ms}$ This is the time when the current reaches $63.2\%$ of its final value.

(b) Formula for current in an RL series circuit is \begin{align*} I(t)&=\frac{\mathcal{E}}{R}(1-e^{-t/\tau}) \\ \\&=\frac{9}{4}\left(1-e^{-\frac{100\times 10^{-6}}{4\times 10^{-3}}}\right)\\ \\&=55.5\quad {\rm mA}\end{align*}
(c) In this part, the unknown is time. To find it, we must solve the equation $I(t)=0.999I_f$ for $t$. Thus, we have \begin{gather*} I=I_f (1-e^{-t/\tau})\\ \\ e^{-t/\tau}=\left(1-\frac{I}{I_f}\right)\end{gather*} Taking the logarithms from both sides gives $-\frac{t}{\tau}=\ln \left(1-\frac{I}{I_f}\right)$ Thus, by arranging and putting the known values into the equation, we obtain \begin{align*} t&=-\tau \ln \left(1-\frac{I}{I_f}\right) \\ \\&=-\tau\ln(1-0.999) \\ \\ &=6.9\tau \end{align*} In other words, it takes $6.9\times 4\,{\rm ms}=27.6\,{\rm ms}$ for the current to reach nearly its final value which is relatively fast.

Problem (4): A 12-V battery is in series with a resistor of $3\,{\rm \Omega}$ and an inductor with unknown inductance.
(a) After a long time, find the current in the circuit.
(b) What is the current after one time constant?
(c) What is the voltage drop across the inductor after one time constant?
(d) Suppose the time constant of this RL circuit is $0.3\,{\rm s}$. Find the inductance of the circuit.

Solution
(a) Recall that the current in the RL series reaches its final or max value after a long time. You can show this by approaching $t$ to infinity in the current formula that finally we get $I_f=\mathcal{E}/R$. Thus, the final current is $I_f=\frac{\mathcal{E}}{R}=\frac{12}{3}=4\quad {\rm A}$
(b) Setting $t=\tau$ in the current RL series formula, we have \begin{align*} I&=I_f(1-e^{-t/\tau})\\ \\ &=4(1-e^{-\tau/\tau})\\ \\&=4(1-e^{-1})\\ \\ &=4\times 0.632 \\ \\&=2.528\quad {\rm A}\end{align*}
(c) From Kirchhoff's voltage rule, we have $\mathcal{E}-\Delta V_R -\Delta V_L=0$. To find $\Delta V_L$, we must determine first the $\Delta V_R$. From the section of Ohm's law problems the voltage drop across the resistor at time $t=\tau$ is found as below \begin{align*}\Delta V_R &=I(t)R\\&=2.528\times 3\\&=7.584\quad {\rm V}\end{align*} Next, substituting it into the above Kirchhoff's rule expression, we get \begin{gather*}\mathcal{E}-\Delta V_R -\Delta V_L=0\\ 12-(7.584)-\Delta V_L=0\\ \Rightarrow \Delta V_L=4.416\quad {\rm V}\end{gather*}
(d) Inductance divided by resistance is defined time constant in a series RL circuit i.e. $\tau=L/R$. Thus, using this formula and solving for inductance we get \begin{align*}L&= R\tau \\ &=3\times 0.3\\&=0.9\quad {\rm H}\end{align*}

Problem (5): In a circuit containing a $35\,\rm V$ power supply, a $45\,\rm \Omega$ resistor, and a solenoid of inductance $1.25\,\rm mH$ with negligible resistance, all connected in a series with an open switch. The switch is suddenly closed.
(a) How long after closing the switch will the current through the circuit reach two-thirds of its maximum value?
(b) How long after closing the switch will the energy stored in the solenoid (inductor) reach two-thirds of its maximum value?

Solution: This problem concerns the growth of current in an R-L circuit after closing a switch. In these cases, the maximum current in the circuit is given by $I=\frac{\mathcal E}{R}$. As you can see, the final current is independent of the inductance because after a very long time, in an R-L circuit, the solenoid (or inductor) behaves like a normal wire.

(a) The current $i$ in an $R-L$ circuit with emf at any moment after closing the switch is given by the following formula $i=I\left(1-e^{-(R/L)t}\right)$ where $I$ is the final current. We must solve the above equation for $t$, which is the only unknown quantity. Taking the natural logarithms of both sides and some manipulations gives the time $t$ in terms of other quantities as below \begin{gather*} e^{-(R/L)t}=1-\frac iI \\\\ \Rightarrow \boxed{t=-\frac LR \ln \left(1-\frac iI\right)}\end{gather*} Substituting $i=\frac 23 I$ into the above, yields \begin{align*} t&=-\frac{1.25\times 10^3}{45}\ln\left(1-\frac{2/3 I}{I}\right) \\\\ &=30.5\,\rm s\end{align*} Therefore, about $30$ seconds after closing the switch, the current in the circuit reaches two-thirds of its final value.

(b) The total energy stored in an inductor with an inductance of $L$ is given by $U=\frac 12 Li^2$. After closing the switch, the current increases from zero to its final value. Thus, if we substitute the current value at any time into the above, we will give the energy stored in the inductor at any instant of time. $U=U_{max}\left(1-e^{(-R/L)t}\right)^2$ where $U_{max}=\frac 12 LI^2$ is the maximum stored energy. Taking the natural logarithms of both sides and rearranging gives the time $t$ in terms of the other given quantities. \begin{align*} t&=-\frac LR \ln \left(1-\left(\frac{U}{U_{max}}\right)^2\right) \\\\ &=-\frac{1.25\times 10^3}{45}\ln\left(1-\left(\frac{2/3 U_{max}}{U_{max}}\right)^2\right) \\\\ &=16.3\,\rm s\end{align*}

Problem (6): In a parallel R-L circuit as below, after the switch is closed a power supply has been designed so that a constant current passes through it.
(a) What will the current through the inductor at any time be?
(b) At what time does the current through the resistor equal to the current through the inductor?

Solution: This is an interesting and challenging question that from its result can be used to answer the AP Physics C questions on this topic.

Contrary to the previous problems, here the inductor is connected with the resistor in a parallel combination. Before solving this problem, knowing a few points are necessary.
(1) In all the electric circuit problems containing an inductor, just after closing a switch, the inductors act as a broken wire and the current through that branch is zero.

(2) In all the electric circuits, each circuit component in parallel with a battery has the same voltage as the power supply. Here, $\mathcal E=V_{ad}=V_{bc}$ where $V_{ad}=iR$ and $V_{bc}=L\frac{di}{dt}$ is the potential difference across the inductor at any time.

(a) By applying voltage and current rules from Kirchhoff's law problems and solving the corresponding equation, we finally reach the following expression for the current at any time in an $R-L$ parallel circuit. $I=I_{max}\left(1-e^{-(R/L)t}\right)$ where $I_{max}=\mathcal E/R$.

As you noticed, this equation is exactly the same as the current equation in an $R-L$ series circuit.

(b) The Kirchhoff's junction rule at point $a$ gives $i=i_1+i_2$ where $i$ is the current through the battery. Assuming a zero internal resistance for the inductor, the equivalent resistor of this circuit would be equal to $R_{eq}=R$. Applying Ohm's law gives us the current through the battery as $i=\frac{\mathcal E}{R_{eq}}=\boxed{\frac{\mathcal E}{R}}$ We are asked to find the moment that the current through the resistor, $i_1$, is equal to the current through the inductor, $i_2$, i.e., $i_1=i_2$.

Substituting this requirement into the expression obtained from the junction rule above and solving for $t$, yields \begin{gather*} i=i_1+i_2 \\\\ i=i_2+i_2 \\\\ \frac{\mathcal E}{R}=2\times \frac{\mathcal E}{R}\left(1-e^{-(R/L)t}\right) \\\\ e^{-(R/L)t}=\frac 12 \\\\ -\frac RL t=\ln 2^{-1} \\\\ \Rightarrow \boxed{t=\frac LR \ln 2}\end{gather*} In the fifth line, we took the natural logarithm of both sides.

There are also some problems with inductance which is related to the RL series circuits.

Author: Dr. Ali Nemati
Date Published: 3/13/2021
Updated: April 26, 2023