RC circuits, a step skipped in solution

In summary: I think.In summary, the textbook says that a source-free RL circuit is created after the switch is opened, but this is not the case. Current flows in a loop and there is no return path.
  • #1
BarelyOtaku
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Homework Statement

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2cdublu.png


This is problem 7.45 in Fundamental of Electric Circuit 5th edition.
2. Homework Equations

First off, I want to go ahead and say that I don't need the solution I already solved it.

I went ahead and used Nodal analysis when Vs = 30. I placed a node voltage right above the 40kΩ wire say ##V_1##.
I then solved for V1 using Voltage division,
##V_1 = \frac{40} {40+20} * 30 = 20## .

Applying nodal analysis,

##\frac{V_1-30}{20} + \frac{V_1}{40} + \frac{V_1-V_o}{10} = 0##.

∴ ##V_o = 20V##

To my understanding Vo is not equal to V1 because there's a resistor between them. However, the problem is that in the textbook solution assumed it WAS and skipped the nodal analysis step.


After solving my nodal equation I realize indeed that Vo is actually equal to V1.
Could anyone explain why? I'd like to know this so I could stop myself from using Nodal analysis for no reason in the final exam.

Please have mercy on me this is my first post
 
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  • #2
Hi BarelyOtaku, Welcome to Physics Forums!

Note that resistors alone do not cause potential drops. It's current flowing through a resistor that causes a potential drop.

In your circuit, once steady state has been achieved "after a long time", the capacitor has reached some constant voltage and charge and current to or from the capacitor is zero. That makes the current through the 10k resistor also zero. As a result there is no potential drop across that resistor: it must have the same potential at both of its ends.
 
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  • #3
Alright thank you very much! Sorry, you replied pretty fast so I didn't expect a reply.
That explanation makes sense but now I've come into the exact same problem but with current xD

alkca0.png


I couldn't grasp the concept of what happens after the switch is open. The book says it's a source-free RL circuit.
However, couldn't the current just flow through the 4Ω resistor and into the RL circuit?
Snapshot.jpg
 
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  • #4
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  • #5
BarelyOtaku said:
Alright thank you very much! Sorry, you guys replied pretty fast so I didn't expect a reply.
That explanation makes sense but now I've come into the exact same problem but with current xD

alkca0.png


I couldn't grasp the concept of what happens after the switch is open. The book says it's a source-free RL circuit.
However, couldn't the current just flow through the 4Ω resistor and into the RL circuit?
View attachment 82283
current has to have a path to ground. Current would not have a path.

If you're confused, write a KCL or KVL at that node.
 
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  • #6
Oooooh, my mind just exploded from comprehending these basic steps.
I was never told and never understood these fundamentals of current and voltage, nor did I find a reference to understand them; now it makes a lot more sense now,
thank you so much guys!
 
  • #7
I just want to clarify, are you supposed to find the DC steady state value, or are you trying to find the step response of the systems?
 
  • #8
Oh I'm supposed to find the inductor current equation for before and after the switch is turn on
 
  • #9
BarelyOtaku said:
Oh I'm supposed to find the inductor current equation for before and after the switch is turn on
Careful. You are asked to calculate the current before and after the source is disconnected from the inductor, which means after the switch is opened.

Are you required to find the final steady-state current, or are you looking for current as an exponential function of time?
 
  • #10
Well from my understanding before the switch is turned on would be the final steady-state, and after the switch is turned on would be the exponential function of time, since it decays as a source-free function.
 
  • #11
NascentOxygen said:
Careful. You are asked to calculate the current before and after the source is disconnected from the inductor, which means after the switch is opened.

Are you required to find the final steady-state current, or are you looking for current as an exponential function of time?

Well from my understanding both.
Before the switch is turned on would be the final steady-state, and after the switch is turned on would be the exponential function of time; since it decays as a source-free function.
 
  • #12
So the answer you gave for both problems was only the steady state value of these circuits, after at least 5 time constants have passed
 
  • #13
donpacino said:
So the answer you gave for both problems was only the steady state value of these circuits, after at least 5 time constants have passed
Yeah I know, I didn't put the whole solution. I was only highlighting the part of the solution I was confused about.
 
  • #14
BarelyOtaku said:
Before the switch is turned on would be the final steady-state,
I was trying to encourage you to use meaningful terms, terms that will be understood by others. If you get the conventions correct at this early stage in your studies, you'll be set for life. Get them wrong, and you'll find it difficult to unlearn later.

We do not see the switch here turn on. Turning a switch on means making a connection, and we are not witness to that happening here.

The switch has already been ON for a long time before t=0 and conditions have reached a steady state. Then at t=0 the source is switched OFF by moving the switch contacts to the position shown in the diagram, and thereafter some time elapses before the inductor branch reaches a new steady state.

There are two steady state conditions. Between the two, there is a period where transient conditions prevail---for the inductor best described as an exponential decay.
 
  • #15
NascentOxygen said:
I was trying to encourage you to use meaningful terms, terms that will be understood by others. If you get the conventions correct at this early stage in your studies, you'll be set for life. Get them wrong, and you'll find it difficult to unlearn later.

We do not see the switch here turn on. Turning a switch on means making a connection, and we are not witness to that happening here.

The switch has already been ON for a long time before t=0 and conditions have reached a steady state. Then at t=0 the source is switched OFF by moving the switch contacts to the position shown in the diagram, and thereafter some time elapses before the inductor branch reaches a new steady state.

There are two steady state conditions. Between the two, there is a period where transient conditions prevail---for the inductor best described as an exponential decay.

I was just referring to the second problem 7.53 where there's a switch. When donpacino asked me the question I assumed he meant the second one since the first one clearly has it's objective defined. Unless I assumed wrong, but I'm pretty sure he was talking about the second one.

Thanks for the clarification, I appreciate as much help as I could get!
 
  • #16
BarelyOtaku said:
I was just referring to the second problem 7.53 where there's a switch. When donpacino asked me the question I assumed he meant the second one since the first one clearly has it's objective defined. Unless I assumed wrong, but I'm pretty sure he was talking about the second one.

Thanks for the clarification, I appreciate as much help as I could get!
I, too, am discussing the second circuit. That's why I spoke of the "inductor". And of the associated switch which is moved from ON to OFF.
 
  • #17
NascentOxygen said:
I, too, am discussing the second circuit. That's why I spoke of the "inductor". And of the associated switch which is moved from ON to OFF.
Ooooh okay I finally get what you mean. As in the switch is on when it's connected and I reversed the terms.

Alright sorry for making it such a hassle xD
 

1. What is an RC circuit?

An RC circuit is an electrical circuit that consists of a resistor (R) and a capacitor (C) connected in series or parallel. It is used to control the flow of current and voltage in a circuit.

2. What is the purpose of skipping the step in a solution for an RC circuit?

The step skipped in a solution for an RC circuit is the initial charging of the capacitor. This step is often skipped because it is assumed that the capacitor has been charged to its full capacity before the circuit is analyzed, making it easier to solve the circuit.

3. How does a capacitor affect the current in an RC circuit?

A capacitor acts as a temporary storage device for electrical energy. When a capacitor is connected to a power source, it charges up and blocks the flow of current. As the capacitor discharges, the current in the circuit increases.

4. What is the time constant of an RC circuit?

The time constant of an RC circuit is the time it takes for the capacitor to charge up to 63.2% of its maximum voltage when connected to a power source. It is calculated by multiplying the resistance (R) by the capacitance (C) in the circuit.

5. How does the time constant affect the behavior of an RC circuit?

The time constant determines how quickly the capacitor charges or discharges in an RC circuit. A longer time constant means a slower charging or discharging process, while a shorter time constant results in a faster charging or discharging process. This affects the overall behavior of the circuit, such as the rate of change of voltage and current.

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