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RL Circuits Problems with Answers

In this tutorial, RL circuits and related formulas are discussed and then some problems with answers are presented which is 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.

An illustration of RL series circuit

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, $e=2.71828$
$\tau$ which 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 $\tau=\frac{L}{R}$. 

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 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 $\Delta V_R=T(t=0)R=0$ just after switch is closed.

(b) In an RL circuit, the final or maximum current is found by 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, time constant in an RL circuit is the time that takes the current reaches $63.2\%$ 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*}

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 times constant has elapsed?
(d) What is the potential drop across the resistance after two times constant?
(e) What is the potential drop across the inductor after two times constant? 

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 time constant of 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^{(-Rt/L)})\\ \\&=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 constant, 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 times constant 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 a 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 Ohm's law, 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 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 an 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*}


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

Author: Ali Nemati
Date Published: 3/13/2021