Switching a RL circuit

In summary, the problem involves a switch that is initially open and then closed at t=0. The question is asking for the current through R1 immediately after the switch is closed, given the values of V, R1, R2, and L. The current through R1 after the switch has been closed for a long time is also asked. The solution to this problem is not provided, but the user has figured it out on their own. The final question is about the current through R1 immediately after the switch is opened, and the user thought it was 0 but that was incorrect.
  • #1
Havoc0040
6
0
1.http://s1191.photobucket.com/albums/z462/havoc0040/?action=view&current=21prob42.jpg
The switch in the figure is open for a very long time and then closed at t = 0. What is the current through R1 immediately after the switch is closed? ( Use V = 3.5 V, R1 = 848 , R2 = 548 , L = 7.2 mH )
After the switch has been closed for a long time what is the current through R1?
After the switch has been closed for a long time what is the current through R2?
 
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  • #2
Hi Havoc0040. Welcome to Physics Forums.

What is your question about this problem? What have you already tried?
 
  • #3
sorry i was able to figure this one out
 
  • #4
gneill said:
Hi Havoc0040. Welcome to Physics Forums.

What is your question about this problem? What have you already tried?

The switch is now opened. What is the current through R11 immediately after the switch is opened? this is my question i thought it was 0 but that's not right
 
  • #5


I would first analyze the given circuit and use Ohm's Law to calculate the initial current through R1 and R2 when the switch is open. This would help in understanding the starting conditions of the circuit before the switch is closed.

Using Ohm's Law, we can calculate the initial current through R1 as I1 = V/R1 = 3.5 V/848 Ω = 0.00413 A. Similarly, the initial current through R2 can be calculated as I2 = V/R2 = 3.5 V/548 Ω = 0.00639 A.

Now, when the switch is closed at t = 0, the circuit becomes a series RL circuit. This means that the current through both R1 and R2 will be the same. However, due to the inductor, the current will not immediately reach its steady-state value.

To calculate the current through R1 immediately after the switch is closed, we need to use the formula I = I0(1-e^(-t/τ)), where I0 is the initial current, t is the time and τ is the time constant of the circuit. The time constant can be calculated as τ = L/R1 = 7.2 mH/848 Ω = 0.00849 s.

Substituting these values, we get I = 0.00413 A(1-e^(-0/0.00849)) = 0.00413 A(1-1) = 0 A. This means that immediately after the switch is closed, there will be no current through R1.

Similarly, the current through R2 can be calculated using the same formula, but with the time constant calculated using R2. So, the current through R2 immediately after the switch is closed will also be 0 A.

However, as time passes, the current through both R1 and R2 will increase and reach their steady-state values. This can be calculated using the formula I = V/R, where V is the voltage and R is the resistance of the respective components. So, after a long time, the current through both R1 and R2 will be equal to V/R1 = 3.5 V/848 Ω = 0.00413 A.
 

1. What is a RL circuit?

A RL circuit is a type of electrical circuit that contains a resistor (R) and an inductor (L). The resistor limits the flow of current, while the inductor stores energy in the form of a magnetic field.

2. How does switching affect a RL circuit?

Switching a RL circuit refers to changing the state of the circuit from on to off or vice versa. This can affect the flow of current and the magnetic field in the inductor, leading to changes in voltage and current in the circuit.

3. What is the difference between opening and closing a RL circuit?

Opening a RL circuit means breaking the circuit by disconnecting one or more components, while closing a RL circuit means completing the circuit by connecting all components. This can have different effects on the flow of current and the magnetic field in the inductor.

4. How does the inductor behave when a RL circuit is switched on or off?

When a RL circuit is switched on, the inductor initially resists the change in current and creates a back EMF (electromotive force) to try and maintain the current. When switched off, the inductor attempts to maintain the current and creates a voltage spike that can damage the circuit.

5. How can switching a RL circuit be used in practical applications?

Switching a RL circuit can be used in various applications such as creating power surges in motors, generating electromagnetic pulses, and regulating current in power supplies. It is also commonly used in electronic switches and relays to control the flow of electricity in circuits.

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