Capacitor and Switch Potential Difference Calculations

In summary, the switch is initially open and the capacitors are uncharged. The applied potential difference is +210V. In part (a), the potential difference between points c and d is +70V. When the switch is closed, the circuit will experience a momentary transient as the capacitors charge or change charge to a new steady-state. In order to understand what happens when the switch is closed, it is necessary to first determine the initial voltages and charges on all the capacitors.
  • #1
burnst14
53
2

Homework Statement


The switch is initially open and the capacitors are uncharged. The applied potential difference Vab = +210V.
a.) What is the potential difference Vcd?
b.) What is the potential difference across each capacitor when the switch is closed?
c.) How much charge flowed through the switch when it was closed?

Homework Equations


I'm thinking this will be all I need:
Q=CV

3. The Attempt at a Solution
I can't even visualize what happens when the switch closes, so I'd like to start there. When the switch closes, more current will divert to the bottom capacitor since it will offer less resistance, but my solution manual says it changes the entire circuit setup.
I've attached the problem statement and the solution manual setup statement.
Screenshot_2015-02-15-14-58-00.png
Screenshot_2015-02-15-14-59-26.png
 
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  • #2
You do realize that this is a DC setup with capacitors as the load, right? So any current that flows will be a momentary transient while the capacitors charge or change charge to a new steady-state. There are no resistances in the circuit.

Have you worked out the initial voltages and charges on all the capacitors prior to the switch closing? It doesn't make sense to start at part (b) if you haven't established the initial conditions for part (b) to take place.
 
  • #3
Okay so, heeding your advice, I solved part (a). I got a potential difference of +70 V from point c to point d.

Now, closing the switch, what happens? I still can't picture it.
 
  • #4
burnst14 said:
Okay so, heeding your advice, I solved part (a). I got a potential difference of +70 V from point c to point d.

Now, closing the switch, what happens? I still can't picture it.
You won't picture it until you've labelled the diagram with the charges on each capacitor. I haven't seen your values for those charges.
 

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by a dielectric material.

2. How do capacitors work?

Capacitors work by accumulating charge on their plates when a voltage is applied. The amount of charge that a capacitor can store is determined by its capacitance, which is measured in farads (F).

3. What is the purpose of a switch in a circuit with capacitors?

A switch is used to control the flow of current in a circuit. In the case of a circuit with capacitors, a switch can be used to connect or disconnect the capacitor from the circuit, allowing for the storage or release of electrical energy.

4. How do you calculate the time constant of a circuit with a capacitor and a switch?

The time constant (τ) of a circuit with a capacitor and a switch can be calculated by multiplying the capacitance (C) of the capacitor by the resistance (R) in the circuit. τ = RC

5. What are some common applications of capacitors and switches?

Capacitors and switches are commonly used in electronic circuits for a variety of purposes, such as filtering out unwanted signals, storing electrical energy, and controlling the flow of current. They are also used in power supplies, audio equipment, and digital electronics.

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