Comprehensive Guide To The Growth and Decay of Current in an LR Circuit: Concepts, Derivations, and Applications
The growth and decay of current in an LR circuit is a fundamental concept in electrical circuits, particularly in understanding how inductors respond to changes in current over time, helping with JEE Main 2025. When a circuit consists of an inductor (L) and a resistor (R), the current does not instantly reach its maximum or drop to zero when a voltage source is applied or removed. Instead, it follows an exponential behaviour due to the inductor's opposition to sudden changes in current. This concept is crucial for analysing transient behaviours in electrical circuits. In this article, we’ll explore the processes of current growth and decay, their mathematical derivations, and real-world applications.
What are LR Circuits?
An LR circuit is an electrical circuit consisting of an inductor (L) and a resistor (R) connected in series or parallel with a voltage source. These circuits are used to study how current changes over time when the circuit is switched on or off. The inductor resists sudden changes in current, leading to transient states of growth and decay of current, which follow an exponential pattern. LR circuits are widely applied in power systems, signal processing, and electromagnetic devices.
What is the Growth and Decay of Current in an LR Circuit?
Growth of Current In LR Circuit: When the circuit is switched on, the current does not reach its maximum value instantly. Instead, it gradually increases, following an exponential pattern.
Decay of Current In LR Circuit: When the circuit is switched off, the current decreases exponentially due to the energy stored in the inductor.
Growth of Current in LR Circuit
An LR Circuit is analysed in three ways. The first one is the initial state, which is present at the instant of closing the switch or opening the switch in the circuit. The second one is the steady state, which appears at any instant after closing or opening the switch. The third one is steady-state, which appears after a long time after closing and opening the switch. Let’s start with the initial and steady states of an LR circuit.
Let us assume a circuit of EMF E has the inductance ’L’ and the resistance ‘R’, as shown in the figure. The voltage drop across the inductor is VL and the voltage drop across the resistor is VR.
Initial Stage of an LR Circuit at t=0
At t = 0, the inductor offers an infinite opposition to the current flow and hence there is no current flow in the circuit at the time of closing the switch. Due to high opposition to the current flow, the voltage is dropped entirely at the inductor and there is no voltage drop across the resistor.
I.e., at t = 0, VR= 0 and VL = E.
Steady State
At a certain point of time, say t = $\infty$, the current in the inductor does not vary with time after closing or opening the switch for a long period. We can see that the current has reached its maximum value and therefore the inductor does not offer any position to the current flow. So, the voltage drop across the inductor becomes zero and the entire voltage drops across a resistor.
At t=$\infty$, VR = E and VL = 0
In this state, the voltage is dropped both across the resistor and the inductor. At any instant t = 0 and t = ∞ is taken for this state. We know that the voltage drop across the inductor is equal to the inductance multiplied by the rate of change in current across the inductor.
I.e.,$ \mathrm{V}_{\mathrm{L}}=L \dfrac{d I}{d t}$
And the voltage across the resistor is given by VR = IR.
In the transient state, when the switch is closed gradually, the current starts increasing across the inductor. Due to the increase in the current, there will be a self-induced EMF in the inductor which opposes the change of the current in the circuit. Let us take an instant at t = t, the current flowing in the circuit is I as shown in the figure.
Transient State of LR Circuit at Time t = t
Let us apply Kirchhoff's voltage law in this circuit.
$\begin{align} &E-V_{R}-V_{L}=0 \\ &E-L \dfrac{d I}{d t}- IR=0 \end{align}$
Therefore, $\mathrm{E}-\mathrm{IR}=L \dfrac{d l}{d t}$
Taking the integration from t = 0 to t = t and I = 0 to I = I:
$\begin{align} &\dfrac{d t}{L}=\dfrac{d I}{E-I R}\\ &\dfrac{1}{L} \int_{0}^{t} d t=\int_{0}^{I} \dfrac{d I}{E-I R}\\ &\dfrac{t}{L}=\left|\dfrac{\ln (E-I R)}{-R}\right|_{0}^{I}\\ &\dfrac{-t R}{L}=(\ln (E-I R)-\ln (E))\\ &\dfrac{-t R}{L}=\ln \dfrac{E-I R}{E}\\&\dfrac{E-I R}{E}=e^{\dfrac{-t R}{L}}\\ &1-\dfrac{I R}{E}=e^{\dfrac{-t R}{L}}\\ &1-e^{\dfrac{-t R}{L}}=\dfrac{I R}{E}\\ &I=\dfrac{E}{R}\left(1-e^{\dfrac{-t R}{L}}\right) \end{align}$
Here, $\dfrac{E}{R}$ becomes the maximum current when there is no inductor opposing the current flow. And $\dfrac{L}{R}$ is called the time constant of the LR circuit represented by $\tau$. It is the time for the current to reach 63% of the final current flowing in the circuit. Therefore, putting these values in the equation, we get the final current equation for the growth of the current in the circuit.
$ I=I_{0}\left(1-e^{\dfrac{-t}{\tau}}\right)$
This is the instantaneous current at time t = t flowing through the circuit.
Let us draw the graph between current and time and see how the current is increasing with time in the growth of the current state.
The Graph between Current and Time in the Growth Stage.
Let us similarly derive the current equation in the decaying state of the current.
Decay of Current
In the Decay of current, the source EMF is removed from the circuit. The current flowing in the circuit will be maximum at the time of this connection of the source. Let maximum current be I0 flowing through the circuit.
Decay of Current
At t = 0, the current flowing in the circuit is I0 and at t = t current flowing is I.
By applying Kirchhoff's voltage law and using integration, we obtain:
$\begin{align} &-L \dfrac{d I}{d t}-I R=0 \\ &-L \dfrac{d I}{d t}=I R \\ &\dfrac{d I}{I}=-\dfrac{R d t}{L} \\ &\int_{I_{0}}^{I} \dfrac{d I}{I}=-\dfrac{R}{L} \int_{0}^{t} d t \\ &\ln (I)_{I_{0}}^{I}=-\dfrac{R t}{L} \\ &\ln \left(\dfrac{I}{I_{0}}\right)=-\dfrac{t}{\tau} \\ &\dfrac{I}{I_{0}}=e^{-\dfrac{t}{\tau}} \\ &I=I_{0} e^{-\dfrac{t}{\tau}} \end{align}$
Therefore, this will be the equation of current decay in the LR circuit.
We derived the LR circuit formula for growth and decay of current in the RL circuit. This LR circuit derivation is very similar in both the cases of growth and decay of current in the LR circuit. Also here, $\dfrac{L}{R}$ is known as the time constant and it is denoted by $\tau$. Let us solve a problem with this concept.
Key Features of Growth and Decay in LR Circuits
Initial State: Att=0, the inductor offers maximum opposition to the current, and the current is zero during growth.
Transient State: The current changes exponentially during this phase.
Steady State: The current reaches its maximum value in growth or becomes zero during decay.
Real-World Applications of LR Circuits
Power Supplies: LR circuits help regulate current changes in power systems.
Signal Processing: Used in filtering signals and smoothing voltage outputs.
Electromagnetic Devices: Essential in designing transformers, relays, and inductors.
Circuit Protection: Decay of current is used to safely dissipate energy in circuit breakers.
Example: In an LR circuit, when the switch is closed and at an instant t = t, the value of current is 0.1 times the value of maximum current. Given that, the time constant is 5 seconds, find it.
Solution: Given: Current at the instant t = t is 3 times the maximum current i.e. I = 0.1I0 and $\tau$ = 5 sec.
We know that when the switch is closed, the current starts increasing in the circuit. The LR circuit formula for current in growth is given by:
$I=I_{0}\left(1-e^{\dfrac{-t}{\tau}}\right)$
Substituting the given values in the above equation, we get:
$\begin{align} &0.1 I_{o}=I_{o}\left(1-e^{\dfrac{-t}{5}}\right) \\ &0.1=\left(1-e^{\dfrac{-t}{5}}\right) \\ &e^{\dfrac{-t}{5}}=0.9 \\ &\dfrac{-t}{5}=\ln (0.9)=-0.105 \\ &t=5 \times 0.105=0.525 \text { seconds } \end{align}$
This is the required value of time needed for the current to become 0.1 times the value of the maximum current.
Conclusion
The LR circuit consists of three stages which are initial, transient and steady states. We have seen how the current in an RL circuit flows with the help of the LR circuit formula and LR circuit derivation. We learned what is the time constant of the LR circuit and how the growth and decay of current in the LR circuit works out.
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FAQs on Growth and Decay of Current in LR Circuits for JEE Main Physics 2025
1. What are the three different stages of current flow and analysis in an LR circuit?
There are three different stages in which the LR Circuit is analysed. These are the initial state, transient state, and steady-state. In the initial state, the current increases across the inductor, and the inductor offers a large opposition to the current flow at the instant of closing the switch. Whereas in the transient state, there is a voltage drop across the inductor and resistor at any instant of time. In the steady state, the current attains its maximum value, and thereby the inductor will not produce any opposition to the current flow. In this steady state, the entire voltage drops across the resistor.
2. What is the weightage of this topic in JEE?
The topic of LR circuits is very important in the JEE examination. Generally in every exam, a minimum of one question from LR circuits, LC circuits, or RC circuits will be asked. This concept is the basis of any wide variety of concepts in AC circuits. The topic of growth and decay of current in the LR circuit and the formulae are very important if we want to find the current flowing in the circuit, consisting of resistors and inductors at a certain point of time, when the switch is closed or opened.
3. What are the stages of current flow in an LR circuit?
The three stages of current flow in an LR circuit are:
Initial state: Current begins to flow, and the inductor offers high opposition.
Transient state: Current changes exponentially, and there’s a voltage drop across both inductor and resistor.
Steady-state: The current reaches maximum, and the inductor stops opposing the current flow.
4. Why is the growth and decay of current in an LR circuit important?
It explains how current varies over time when the circuit is switched on or off. This concept helps in calculating current at any instant and analysing transient behaviours in circuits with inductors and resistors.
5. What is the weightage of the LR circuit topic in JEE?
The topic holds significant importance in JEE, with at least one question frequently appearing from Growth and Decay of Current in LR Circuit, LC circuits, or RC circuits. Its formulas are critical for solving problems related to current flow at specific times.
6. How does the Growth Of Current In LR Circuit happen?
When a switch is closed in an LR circuit, the current gradually increases due to the inductor's opposition to sudden changes. The growth of current follows an exponential pattern described by:
$I(t) = I_{max}(1 - e^{-t/\tau})$
where $\tau$ is the time constant.
7. How does the decay of current occur in an LR circuit?
When the circuit is opened, the current decreases exponentially due to the inductor's stored energy. The decay is expressed as:
$I(t) = I_{max}e^{-t/\tau}$
This describes how the current fades away with time.
8. How can we increase the current in an LR circuit?
To increase current in an LR circuit:
Reduce the resistance R.
Increase the supply voltage V.
Use an inductor with lower inductance L to minimise opposition during transient states.
9. What is the significance of the time constant $\tau$ in an LR circuit?
The time constant$\tau = L$ determines the rate of current growth and decay. A larger $\tau$ means slower current changes, while a smaller $\tau$ leads to quicker adjustments.
10. Where can I find Growth and Decay of Current in LR Circuit PDFs for preparation?
PDFs explaining the Growth and Decay of Current in the LR Circuit are available on our Vedantu. They include detailed explanations, derivations, and solved problems.
11. How is voltage distributed during the transient state in an LR circuit?
In the transient state, the voltage is divided between the inductor and resistor. The inductor's voltage gradually decreases as the current increases, while the resistor's voltage increases until steady state is reached.
12. What happens to the inductor in the steady state of an LR circuit?
In the steady state, the inductor acts as a short circuit, producing no opposition to current flow. All the voltage drops across the resistor.
13. What is the role of inductance in the growth and decay of current?
Inductance L controls how quickly or slowly the current changes during growth and decay. A higher L causes slower current changes.
14. How do LR circuits apply to real-world systems?
LR circuits are used in devices like transformers, power supplies, and motor controls to manage current changes and protect circuits during transient states.
15. Can the growth and decay of current be observed experimentally?
Yes, the exponential behaviour of growth and decay in an LR circuit can be measured using oscilloscopes or circuit simulation software.
16. Why are LR circuits important for AC analysis?
LR circuits form the basis of AC circuit analysis, helping understand inductive reactance, resonance, and power factor corrections.
17. How is the concept of decay of current applied in circuit breakers?
The decay of current in LR circuits helps dissipate stored energy in inductors, preventing voltage spikes and ensuring safe circuit interruption.
18. How To Increase Current In A Circuit?
To increase current in a circuit:
Reduce resistance by using lower-resistance materials or components.
Increase the voltage of the power supply to drive more current.
Ensure all connections are proper to avoid unnecessary resistance.
Use devices with low inductance in circuits with inductors to reduce opposition to current changes.
