One of my lab exercise is to build a circuit with 12V DC supply to 3 capacitors: 0.15 MicroF, 0.25 Micro F and 0.35 MicroF in series. Then measure voltage across each capacitor. But there is a note that we need to discharge all capacitors again if making a mistake when doing voltage measurement across them. Could someone explain why this is necessary and important? Thank you!

 

But there is a note that we need to discharge all capacitors again if making a mistake when doing voltage measurement across them. Could someone explain why this is necessary and important?

Likely because the mistake could have caused some discharge of one or more of the capacitors, affecting their voltage.
So you discharge all the capacitors and start over to make sure the measurements are accurate.

Did the instructor ask you to calculate what the expected cap voltages should be?

 

Yes, they did ask to calculate the voltage drop across the caps and compare with the measured values. Could you explain in more details? If they are in circuit with voltage supply then they would be charged again? Right?
Thanks.

 

Could you explain in more details? If they are in circuit with voltage supply then they would be charged again? Right?

Yes.
You need to discharge all the caps at the start so there is no residual charge to affect the final value.
Then, when you apply the voltage to the 3 caps in series, the voltage/charge will distribute in inverse proportion to the capacitor values.
Here's a discussion on capacitors in series.

 

One of my lab exercise is to build a circuit with 12V DC supply to 3 capacitors: 0.15 MicroF, 0.25 Micro F and 0.35 MicroF in series. Then measure voltage across each capacitor. But there is a note that we need to discharge all capacitors again if making a mistake when doing voltage measurement across them. Could someone explain why this is necessary and important? Thank you!

What instrument are you going to use to measure the voltage across the capacitors? A typical voltmeter you might use could have an input resistance of 10 megohms. How fast will the voltage across a .15 μF change when you connect such a voltmeter across it? Will the speed with the capacitor is discharged by the voltmeter make it difficult to get an accurate measurement?

 

What instrument are you going to use to measure the voltage across the capacitors? A typical voltmeter you might use could have an input resistance of 10 megohms. How fast will the voltage across a .15 μF change when you connect such a voltmeter across it? Will the speed with the capacitor is discharged by the voltmeter make it difficult to get an accurate measurement?

I used a Fluke DMM 789 to measure the voltages. It charge quite fast in few minutes. Yes, it is difficult to measure and you need to get the Max value from beginning of the measurement period. If you want, I could upload the test results here soon.

 

Yes.
You need to discharge all the caps at the start so there is no residual charge to affect the final value.
Then, when you apply the voltage to the 3 caps in series, the voltage/charge will distribute in inverse proportion to the capacitor values.
Here's a discussion on capacitors in series.

Thanks. After reading the thread, I found this interesting detail that may be the answer: "Capacitors connected in series encounter a problem of exceeding their maximum voltage rating, if one capacitor is leaky or results in short circuiting it will cause an over-voltage to its neighbouring capacitor, i.e, exceeding the capacitor’s maximum voltage rating. Also a low value capacitor will eventually experience large amounts of current flow thus exceeding its maximum voltage rating again. Under these extremes “balancing resistors” are required for rectifying these situations".

 

Do you understand the purpose of the exercise? What voltages would you expect on the 3 capacitors, and why?

 

Do you understand the purpose of the exercise? What voltages would you expect on the 3 capacitors, and why?

Yes, the purpose is to explore the role a capacitor play in a circuit, in series and parallel later and also try to find the RC time constant.

 

I assume you used the formula Q = C*V to determine the voltage across each capacitor.

If the capacitors are all fully discharged when you connect them to the battery, then each capacitor will receive the same charge Q = .93 μC. Knowing this, you can calculate the expected voltage across each capacitor using Q = C*V.

Suppose one capacitor already had some charge. Then when you connect the battery the increment of charge added to the three capacitors will be the same for each capacitor, but the one capacitor that already had some charge will have more than the increment added by connecting to the battery. How will this affect the voltage across that capacitor that started with some charge? Or, suppose the three capacitors had different amounts of charge already in them when you connect the battery; in that case, will each capacitor have the same charge after connecting the battery?

Can you think of a way that the capacitors might have different amounts of charge remaining after you measure their voltages with your meter?

 

I thought the new charge would be distributed evenly among 3 caps so that the final charge Q=Q1=Q2=Q3 accordimg to theory, charges of caps in series are equal. However, I will do it again in my lab today to see how it works. Thanks for your explanation!

 

I thought the new charge would be distributed evenly among 3 caps so that the final charge Q=Q1=Q2=Q3 accordimg to theory, charges of caps in series are equal. However, I will do it again in my lab today to see how it works. Thanks for your explanation!

The new charge would be distributed evenly, but what if there were already some charge on one cap? Or what if there were an already existing uneven distribution of charge before the new charge was added?

Do you see how connecting the voltmeter to the caps could drain more charge from one cap than from the others? What if you left the voltmeter connected to one cap longer than to the others? And even if the voltmeter were connected to each cap exactly the same length of time, would more charge be drained from the cap with the highest voltage to start with?