Initial Voltage in an RC circuit with a Current Source and no Voltage Source

In summary, the conversation is discussing a circuit problem involving a resistor and a capacitor. The problem involves finding the initial conditions of the circuit, specifically the voltage across the capacitor at t=0. The conversation includes a screenshot of the circuit and equations for V=IR and V=V0exp(-t/tau). The discussion centers around the concept of short circuiting and how it affects the flow of current in the circuit. The conclusion is that at t=0, the current will not flow through the resistor and therefore there will be no voltage drop across it. This results in the initial voltage across the capacitor being 0V.
  • #1
Bassa
46
1

Homework Statement


[/B]
Derive an expression for vc(t) in the circuit of Figure P4.13 and sketch vc(t) to scale versus. I attached a screen shot of the circuit for this problem.

Screen Shot 2017-02-19 at 12.04.57 AM.png
time.

Homework Equations



V = IR
V = V0exp(-t/tau)

The Attempt at a Solution


[/B]
I am trying to find the initial conditions to this problem. To do this, I have said that at t < 0 the capacitor will act as a short circuit for a dc current. we would then have that the current that is going through the resistor R is 10mA. This would yield to Vc(0) = (.01)(2000) = 20V. When I checked a solution of this problem, they get that Vc(0) = 0V. I am not really sure why that is! Thank you very much for your help!
 
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  • #2
Bassa said:

Homework Statement


[/B]
Derive an expression for vc(t) in the circuit of Figure P4.13 and sketch vc(t) to scale versus. I attached a screen shot of the circuit for this problem.

View attachment 113412 time.

Homework Equations



V = IR
V = V0exp(-t/tau)

The Attempt at a Solution


[/B]
I am trying to find the initial conditions to this problem. To do this, I have said that at t < 0 the capacitor will act as a short circuit for a dc current. we would then have that the current that is going through the resistor R is 10mA. This would yield to Vc(0) = (.01)(2000) = 20V. When I checked a solution of this problem, they get that Vc(0) = 0V. I am not really sure why that is! Thank you very much for your help!
Short circuit means zero resistance. Would the current flow through the resistor, when it can choose a parallel zero-resistance path?
 
  • #3
ehild said:
Short circuit means zero resistance. Would the current flow through the resistor, when it can choose a parallel zero-resistance path?

Thanks you for the quick reply! Oh! I I did not think of it this way. My answer would be "No, it would not." That raises another question that might be related to a fundamental piece of information that I don't know. Current would also not flow if there was no change in voltage, which the resistor offers, right?
 
  • #4
Bassa said:
Thanks you for the quick reply! Oh! I I did not think of it this way. My answer would be "No, it would not." That raises another question that might be related to a fundamental piece of information that I don't know. Current would also not flow if there was no change in voltage, which the resistor offers, right?
Yes, no current through the resistor implies zero voltage across it.
But you can think in the point of view of the capacitor. At t=0, there is zero charge on it. The voltage across a capacitor is V=Q/C. So zero charge means zero voltage across the capacitor, and the resistor and capacitor are parallel connected, so V(t)=0 at t=0.
Is your "relevant equation" V = V0exp(-t/tau) correct then?
 
  • #5
ehild said:
Yes, no current through the resistor implies zero voltage across it.
But you can think in the point of view of the capacitor. At t=0, there is zero charge on it. The voltage across a capacitor is V=Q/C. So zero charge means zero voltage across the capacitor, and the resistor and capacitor are parallel connected, so V(t)=0 at t=0.
Is your "relevant equation" V = V0exp(-t/tau) correct then?

So, I think that I am misunderstanding the problem. The arrow on the switch in the diagram means that the switch was closed for a long time and then it was opened at t=0s. If this is the case, then would that not mean that the capacitor was charged at t <= 0?
 
  • #6
Bassa said:
So, I think that I am misunderstanding the problem. The arrow on the switch in the diagram means that the switch was closed for a long time and then it was opened at t=0s. If this is the case, then would that not mean that the capacitor was charged at t <= 0?
If the switch is closed, it means short-circuit for the current source. Why should the current choose a different path?
 
  • #7
ehild said:
If the switch is closed, it means short-circuit for the current source. Why should the current choose a different path?

I think I am starting to understand where you are coming from. Initially, I thought that current would not go through the parallel piece of wire because there is no voltage drop there, but it would go to the resistor because there there is a voltage drop and the current could flow. I have always thought that current would not go throw a wire that did not have a voltage drop. I think this is where my misunderstanding is coming from.
 
  • #8
Moderator's note: Changed the thread title to make it less generic and more specific to the problem being addressed.
 
  • #9
gneill said:
Moderator's note: Changed the thread title to make it less generic and more specific to the problem being addressed.

Thank you for the note. I changed the title of the thread.
 
  • #10
ehild said:
If the switch is closed, it means short-circuit for the current source. Why should the current choose a different path?

Would the current not choose the path that has a voltage drop, which is the one with the resistor?
 
  • #11
Bassa said:
Would the current not choose the path that has a voltage drop, which is the one with the resistor?
Current *causes* a voltage drop when it flows through a resistor. If it's flowing elsewhere, there will be no voltage drop.
 
  • #12
Bassa said:
Would the current not choose the path that has a voltage drop, which is the one with the resistor?
How do you know that there is a voltage drop across the resistor? The voltage across a resistor is U=IR. The resistor is not connected to a voltage source, the voltage across it is determined by the current. Zero current, zero voltage.
The branch including the closed switch has zero resistance. So the voltage across the current source is zero. So is the voltage across both the resistor and capacitor.
 
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  • #13
ehild said:
How do you know that there is a voltage drop across the resistor? The voltage across a resistor is U=IR. The resistor is not connected to a voltage source, the voltage across it is determined by the current. Zero current, zero voltage.
The branch including the closed switch has zero resistance. So the voltage across the current source is zero. So is the voltage across both the resistor and capacitor.

Thank you very much for this explanation! I really appreciate it!
 
  • #14
gneill said:
Current *causes* a voltage drop when it flows through a resistor. If it's flowing elsewhere, there will be no voltage drop.

Thank you!
 

1. What is the significance of initial voltage in an RC circuit?

The initial voltage in an RC circuit refers to the voltage that is present in the circuit before any changes or inputs are made. It is important because it affects the behavior of the circuit and can impact the flow of current.

2. How is the initial voltage calculated in an RC circuit with a current source?

The initial voltage in an RC circuit with a current source is calculated using Ohm's law, which states that voltage is equal to current multiplied by resistance. In this case, the initial voltage is equal to the current source value multiplied by the resistance in the circuit.

3. What happens to the initial voltage when a voltage source is added to the RC circuit?

When a voltage source is added to an RC circuit, the initial voltage may change depending on the value of the voltage source. If the voltage source is greater than the initial voltage, the initial voltage will be overridden. If the voltage source is less than the initial voltage, the initial voltage will remain and affect the flow of current in the circuit.

4. Can the initial voltage change over time in an RC circuit with a current source?

In an RC circuit with a current source and no voltage source, the initial voltage will remain constant as long as the current source remains constant. However, if there are changes in the circuit such as adding or removing components, the initial voltage may change accordingly.

5. How does the initial voltage affect the charging and discharging of the capacitor in an RC circuit?

The initial voltage affects the charging and discharging of the capacitor in an RC circuit by determining the starting point of the voltage on the capacitor. If the initial voltage is high, the capacitor will take longer to charge and vice versa. It also affects the rate of change of the capacitor's voltage during discharging.

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