Initially, charge is shared between Source C1 and the two Co capacitors which make up Ctotal. Once charge is stabilized, the wire link is cut, forming a tiny capacitor Cm in series which is added to C total, reducing Ctotal to zero. What happens to the charges and voltages in the loop then, and also when the gap Cm is opened and closed repeatedly?
Electronic – What happens to charge in a 3-capacitor loop when link between two capacitors is opened and closed
capacitorchargeseries
Related Solutions
The key is to think about what happens to the gain at DC when you couple with the capacitors (this is assuming the capacitors are large enough not to attentuate the signal passed through significantly - I'm assuming this probably isn't the point of the question since the values are not shown)
If you apply no signal to the input in each scenario, what is the output voltage?
The basic rule is you can't instantly change the voltage across a capacitor.
Q1 is a switch operated by an input to Q2 so redraw the circuit using a (controlled) switch which makes it easier to see what's happening.
When the circuit is powered up C1 an C2 charge up to about 2V. C2 will take longer to charge than C1 because it has R1,R2 and R5 reducing its charging current. After charging there is no current through R5 so there is no voltage difference between its ends. i.e. we now have 0V at the end of both capacitors and +2V at their positive ends. (initial charging circuit)
When Q1,Q2 is turned on the top end of R5 immediately jumps to +2V (held there by the voltage on C1).
Following the rule that we cannot immediately change the voltage across a capacitor then we must still have 2V across C2. This means that the positive end of C2 must be immediately raised to +4V because the other terminal is at +2V.
N.B. the rule does not say we can't instantly change the voltage at the capacitors terminals only that the voltage difference between them cannot change immediately. The rule is a consequence of the principle of charge conservation.
Neither capacitor has been discharged at this point (charge has been conserved) but we have effectively doubled the voltage at the output.
Best Answer
Once the capacitors are all charged and no current is flowing in them the cicuit you get is this one:
simulate this circuit – Schematic created using CircuitLab
disconnect the wire between the two capacitors Co and the circuit you get looks like this
simulate this circuit
Since VCo=VC1/2 we now have a voltage between the terminals (split wire ends) of -VCo+VC1-VCo=0v
Now you can of course reconnect the two wire ends, but as there is no voltage difference between them no current will flow.
As Neil sayed a lot of nothing happens.
HOWEVER;
If you were to actually change the capacitance of a capacitor while it is charged what would happen is that the voltage would change. considder this schenario;
You have a capacitor (two plates opposite each other) charged to some voltage. now you move them appart by some distance, the capacity is going to decrease but because there is the same amount of energy stored in the capacitor the voltage will increase to compensate so that Q=(V^2*C)/2 stayes unchanged.