# Analysis of an RL Circuit: Voltage, Current and Time Dependence

• Mola
In summary, Momodou is asking for clarification on their analysis of an RL circuit. They believe that after the switch is closed, the current in R1 is at its maximum due to being in parallel with R2, which has a current of zero. They also believe that as time goes to infinity, the current in R1 will decrease to zero while the current in R2 will reach equilibrium. However, the current in R1 does not actually decay to zero as it is still connected to the battery. The inductor acts as a straight wire, causing the current in each branch to be obtained by considering R1 and R2 in parallel with the battery. Momodou is seeking help in understanding this interesting circuit.

#### Mola

I am trying to analyse an RL circuit, particulartly the dependence of current in 2 resistors with time, in an RL circuit. This is what I think but I would really love some people to tell me if I am getting it wrong or if there is anything they want to add to it.

***Just after the the switched at t = 0, the voltage in R1 is the same as the voltage in R2 because they are in parallel. Therefore the current in R1 is: I = E/R1 which is the maximum current. But the current in R2 is zero because current in the inductor is zero at t = 0.

Question:
Do I have a good reasoning for the current in the resistor R2 being zero at t = 0?

***A long time after the switched is closed: I think as time goes to infiniti, the current in R1 drops down from maximum (E/R1) to zero.
The current in R2 starts growing exponentially as time increases. And at some point the current in R2 goes steady or reach equilibrium(not growing and not decaying).

Question: Is it true that the current in R1 decays to zero as time goes to infiniti? I am not sure the current in R1 decays though because the resistor does not consume current.

I would like some help analysing this circuit. It looks pretty interesting.

Thanks, Momodou

#### Attachments

• drawing1.bmp
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The answer to the first question is yes, your reasoning is correct.
The answer to the question is no. The current in R1 does not decay to zero because it is still connected across the battery. A long time after the switch is closed, the current in the branch with the inductor is not changing. This means that the inductor acts as a straight wire (short). The current in each branch is obtained by considering R1 and R2 in parallel with the battery.
(I changed the original figure format from BMP to PNG and uploaded it in case anyone looks at this post)

## 1. What is an RL circuit?

An RL circuit is a type of electrical circuit that contains a resistor (R) and an inductor (L). It is used to study the behavior of voltage, current, and time in response to changes in the circuit.

## 2. What is the time constant of an RL circuit?

The time constant of an RL circuit is a measure of how quickly the current in the circuit reaches its maximum value. It is calculated by multiplying the resistance (R) by the inductance (L), and is denoted by the symbol τ (tau).

## 3. How does the voltage behave in an RL circuit?

The voltage in an RL circuit is initially high and then decreases over time as the current in the circuit increases. This is due to the presence of the inductor, which resists changes in current and causes a voltage drop.

## 4. What is the relationship between current and time in an RL circuit?

The current in an RL circuit follows an exponential decay curve, where it starts at its maximum value and decreases over time. The rate of decrease depends on the time constant, with a shorter time constant resulting in a faster decrease in current.

## 5. How can the behavior of an RL circuit be predicted?

The behavior of an RL circuit can be predicted by using mathematical equations and formulas, such as Ohm's law and the equation for the time constant. Additionally, simulations and experiments can also be performed to observe the behavior of the circuit in real-life situations.