Solving an RC Circuit Using Differential Equations

In summary, the conversation discusses two different methods for finding the current in a circuit after a switch is opened. Method 1 involves calculating the voltage at the open circuit, while method 2 involves considering the conditions at t=0- and t=0+. The conversation also addresses a question about why Voc is equal to 12, explaining that there is no current or IR drop in the resistors, resulting in a voltage of 12 at the -ve side of Voc.
  • #1
theBEAST
364
0

Homework Statement


hwAvh.png


The Attempt at a Solution


I tried two different methods and came to an odd conclusion:
qEbub.png
 
Physics news on Phys.org
  • #2
The situation begins with the switch closed.
Before the switch is opened, the capacitor will charge up to some voltage, driven by both power supplies. You can calculate what that is.

When the switch opens, it removes one of the voltage sources so that the capacitor begins to discharge into the resistor network. A dropping current will start to flow and you are asked to find what that current is 0.1 seconds after the switch is opened.
 
  • #3
In method 1, Voc is not zero. No current flows so Voc = 12. If Voc were zero, you would be claiming the open circuit behaved like a short circuit and there would be a path for current to flow through Voc to the resistors.

In method 2, you are mixing up the conditions at t=0- and t=0+. At t=0-, the switch is closed and the capacitor is open (no current flows through it). At t=0+, the switch is open but the capacitor is in there because current will flow through it. The connection between t=0- and t=0+ is that currents and voltages cannot instantly change so the initial condition prior to the switch opening will be the same as after it opens.
 
Last edited:
  • #4
aralbrec said:
In method 1, Voc is not zero. No current flows so Voc = 12. If Voc were zero, you would be claiming the open circuit behaved like a short circuit and there would be a path for current to flow through Voc to the resistors.

In method 2, you are mixing up the conditions at t=0- and t=0+. At t=0-, the switch is closed and the capacitor is open (no current flows through it). At t=0+, the switch is open but the capacitor is in there because current will flow through it. The connection between t=0- and t=0+ is that currents and voltages cannot instantly change so the initial condition prior to the switch opening will be the same as after it opens.

Wait, why is Voc = 12? If there's no current through the three resistors on the left loop then the voltage would be zero wouldn't it?
 
  • #5
theBEAST said:
Wait, why is Voc = 12? If there's no current through the three resistors on the left loop then the voltage would be zero wouldn't it?

No, there is no current in the three resistors so there is no IR drop either so the voltage at the -ve terminal of the battery is the same as at the -ve side of Voc. You can make the same argument for R5 and come to Voc= 12.
 

Related to Solving an RC Circuit Using Differential Equations

What is an RC circuit?

An RC circuit is a type of circuit that contains a resistor (R) and a capacitor (C) connected in series or parallel. It is commonly used in electronic devices to filter or shape electrical signals.

Why do we use differential equations to solve RC circuits?

Differential equations are used to describe the behavior of dynamic systems, such as RC circuits. They take into account the changing values of voltage and current over time, making them an effective tool for analyzing and solving complex circuits.

What are the steps for solving an RC circuit using differential equations?

The steps for solving an RC circuit using differential equations are as follows:

1. Write the differential equation for the circuit using Kirchhoff's voltage law and Ohm's law.

2. Simplify the equation by substituting any known values and rearranging terms.

3. Solve the differential equation using integration to find the general solution.

4. Apply initial conditions, such as the initial voltage or current values, to find the particular solution.

5. Use the particular solution to find the voltage or current at any given time.

What are the applications of solving RC circuits using differential equations?

Solving RC circuits using differential equations is important in the design and analysis of electronic circuits. It can help engineers predict the behavior of a circuit and make necessary adjustments to optimize its performance. It is also used in fields such as telecommunications, control systems, and signal processing.

Can you use other methods besides differential equations to solve an RC circuit?

Yes, there are other methods for solving RC circuits, such as using phasors, Laplace transforms, or computer simulations. However, differential equations are considered the most accurate and versatile method, and are commonly used in engineering and scientific applications.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
212
  • Introductory Physics Homework Help
Replies
4
Views
1K
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
4K
Back
Top