R-L Circuit, Finding the current through a circuit element.

In summary, the current through the 20 ohm resistor is zero, and the current through the inductor is also zero. This happens because the voltage across the inductor is zero, and the current through the inductor is connected in parallel with the current through the 20 ohm resistor.
  • #1
hyddro
74
2

Homework Statement


So I am having a hard time trying figuring this out. The question asks for the current through the 20Ω resistor as well as the current through the battery AFTER THE SWITCH HAS BEEN CLOSE FOR ALONG TIME.

Here is a picture of the circuit.

http://i.imgur.com/VjoFSUd.jpg

Homework Equations



Kirchhoff Loop rule.
Kirchhoff node rule.

V_L = L*di/dt

V=iR.

i = I (e^(-t(R/L)))

The Attempt at a Solution



Here is how I tried figuring it out. After a long time di/dt =0 so the potential across the inductor is 0. If that is the case then by Kirchhoff Loop rule (outer loop) E - 10Ω*I - V_L = 0 or 30 - 10Ω*I - V_L = 0
since V_L = 0, then the current through the resistor (and the battery) would be 3A. Now, when I appply the loop rule on the right loop. V_L - 20Ω(I) = 0. Again, Since V_L = 0 then the current through the 20Ω resistor must be zero, but this doesn't make any sense to me, i just don't know why but I feel this is wrong. Can anyone confirm this? Also, if this is right, why is the current through that resistor 0 ? Thank you
 
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  • #2
hyddro said:

Homework Statement


So I am having a hard time trying figuring this out. The question asks for the current through the 20Ω resistor as well as the current through the battery AFTER THE SWITCH HAS BEEN CLOSE FOR ALONG TIME.

Here is a picture of the circuit.

http://i.imgur.com/VjoFSUd.jpg

Homework Equations



Kirchhoff Loop rule.
Kirchhoff node rule.

V_L = L*di/dt

V=iR.

i = I (e^(-t(R/L)))

The Attempt at a Solution



Here is how I tried figuring it out. After a long time di/dt =0 so the potential across the inductor is 0. If that is the case then by Kirchhoff Loop rule (outer loop) E - 10Ω*I - V_L = 0 or 30 - 10Ω*I - V_L = 0
since V_L = 0, then the current through the resistor (and the battery) would be 3A. Now, when I appply the loop rule on the right loop. V_L - 20Ω(I) = 0. Again, Since V_L = 0 then the current through the 20Ω resistor must be zero, but this doesn't make any sense to me, i just don't know why but I feel this is wrong. Can anyone confirm this? Also, if this is right, why is the current through that resistor 0 ? Thank you

You just figured out that the voltage across the inductor is zero, and the 20 ohm is across the inductor also. So what would be the voltage across the 20 ohm resistor? the current?
 
  • #3
Oh Yeah! They are connected in parallel, so the Potential across the inductor must be equal to the potential across the 20 ohm resistor, if that is the case then the potential across the resistor is 0, from Ohm's law, 0=I.R, so I must be zero as well... Is this correct? Why is the current 0? I mean mathematically this makes sense, but in terms of the role of an inductor in a circut, why is this happening? What happens if we replace this inductor with a piece of wire and close the switch? would this scenario still happen? Thank you!
 
  • #4
Well, why shouldn't the current thru the 20 ohm be zero?

If you let current in an iinductor settle at a constant value - any constant vaue including zero - then the voltage across that inductor is zero.

If you replaced the inductor with a wire then obviously the current thru the 20 ohm would always be zero.

Ashort wire is an inductor with a very low inductance (a few nanohenries typically).
 
  • #5
ok that was really helpful. Thank you Rude man! hehe
 
  • #6
Edit...
 

What is an R-L circuit?

An R-L circuit is a type of electrical circuit that contains both resistors (R) and inductors (L). These components work together to regulate the flow of electrical current through the circuit.

How do you find the current through a circuit element in an R-L circuit?

The current through a circuit element in an R-L circuit can be found using the formula I = V/R, where I is the current in amperes, V is the voltage in volts, and R is the resistance in ohms. This formula can be modified to include the inductance of the circuit element, which would result in a more accurate calculation.

What is the purpose of an inductor in an R-L circuit?

An inductor is used in an R-L circuit to store energy in the form of a magnetic field. This magnetic field opposes changes in the current flowing through the circuit, helping to regulate and stabilize the flow of electricity.

How does the current change over time in an R-L circuit?

In an R-L circuit, the current changes over time in a characteristic pattern known as an exponential decay. This means that the current decreases rapidly at first and then levels off over time.

What factors can affect the current in an R-L circuit?

The current in an R-L circuit can be affected by various factors, including the voltage supplied, the resistance of the circuit, the inductance of the inductor, and the frequency of the current. Additionally, the presence of other components in the circuit, such as capacitors, can also impact the current flow.

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