Voltage Measurement in a Capacitor Circuit with AC Charging

[John4]

Homework Statement

Hi

I have been told that the question below might be similar to potential university interview questions for physics/electronic engineering.

In relation to this circuit diagram:

http://forum.allaboutcircuits.com/attachments/diagram-ac-diode-capacitor-pd-question-png.90713/

Initially 240V rms mains voltage is connected between terminals A and B in the circuit above. You can assume that the capacitors and diodes are appropriately rated to work at this voltage.
When the capacitors have ceased to charge, the mains voltage is disconnected.
What voltage will then be measured between points C and D?
To my mind the voltage between C and D will just be the sum of the voltages across the two capacitors but C1 will not charge so the 240V rms will be all across C2. This will result in the voltage between C and D to be 240V.

Homework Equations

The Attempt at a Solution

To my mind the voltage between C and D will just be the sum of the voltages across the two capacitors but C1 will not charge so the 240V rms will be all across C2. This will result in the voltage between C and D to be 240V.[/B]

 

[jackMybrain@ru]

Hi, are you working with an AC voltage source?? Because you are putting only 240V rms in AC circuit and there are capacitor which also consume imaginary voltage from an angular frequency. From your analogy, you are considering only the A port and thinking it to solve as DC circuit (one way to shortest route). Unfortunately, I am an first year, therefore its quite complex for me, but I think C1 will also provide energy to the circuit while getting charged through B port source. Therefore, you should consider both of the capacitors.

 

[CWatters]

To my mind the voltage between C and D will just be the sum of the voltages across the two capacitors

Correct.

but C1 will not charge so the 240V rms will be all across C2.

That's incorrect. The input is AC and the circuit is effectively symmetrical. When C1 is charging C2 won't be but what happens in the next half of the cycle?

There are several ways to visualise what's going on but you might find it easier to redraw the circuit and imagine A is connected to 0V and B has 240V ACrms applied.

Note it's 240V ACrms not 240V AC pk.

 

[ehild]

The circuit can be redrawn to show the symmetry. How does the current flow when B is positive with respect to A and how, when B is negative with respect to A?

 

[John4]

Thanks for your help.

So am I correct in saying that each capacitor will charge to sqrt(2) * Vrms hence the DC voltage across each cap will be approximately 339V / the voltage from points C to D will be 679V?

 

[ehild]

Yes.

 

[John4]

Thanks again for your help; much appreciated.