Quick question about make before break switches and capacitors

[kl055]

Homework Statement

A capacitor is initially connected to a voltage source in series for a long time. At time t0 the make before break switch switches to another series circuit which includes the capacitor. This means that the capacitor will have an initial voltage v(0-) = voltage source, correct? And if the switch was a break before make switch, the initial voltage v(0-) across the capacitor would be zero, correct?

Homework Equations

The Attempt at a Solution

 

[gneill]

Homework Statement

A capacitor is initially connected to a voltage source in series for a long time. At time t0 the make before break switch switches to another series circuit which includes the capacitor. This means that the capacitor will have an initial voltage v(0-) = voltage source, correct? And if the switch was a break before make switch, the initial voltage v(0-) across the capacitor would be zero, correct?

It would depend upon the details of the circuit and what the switch is accomplishing when it commutes. A diagram would help.

It might help you to know that if a charged ideal capacitor is removed from a circuit (so that one or both of its leads are no longer connected so that there's no closed circuit for current to flow into or out of the capacitor), then it will retain the charge it had at the moment of disconnection indefinitely.

 

[rude man]

Homework Statement

A capacitor is initially connected to a voltage source in series for a long time. At time t0 the make before break switch switches to another series circuit which includes the capacitor. This means that the capacitor will have an initial voltage v(0-) = voltage source, correct? And if the switch was a break before make switch, the initial voltage v(0-) across the capacitor would be zero, correct?

The answer depends on the circuit the capacitor is switched to. Assuming it's not switched to a second hard voltage source, the capacitor voltage would be the same before and after the switching regardless of whether the switch is mbb or bbm: v(0+) = v(0-). That's because a capacitor can't instantly change its voltage unless switched to another hard voltage.

If it's switched to a second hard voltage source a mbb switch would cause instant chaos since you'd be connecting two hard voltage sources to each other. If it's bbm then v(0-) would be the first hard voltage and v(0+) = second hard voltage.

 

[kl055]

Thanks for the replies. I took my electric circuits course a long time ago and I need a refresher. I think I got it now.

 

[rezeew]

Yes, you are correct. In a make before break switch, the capacitor will have an initial voltage equal to that of the voltage source since they are connected in series for a long time before the switch is changed. This is because the capacitor charges up to the same voltage as the source. In a break before make switch, the initial voltage across the capacitor would be zero since the connection to the voltage source is broken before the new connection is made. This means that the capacitor has no charge and therefore no voltage. It is important to consider the type of switch being used when analyzing circuits involving capacitors to ensure accurate results.