| Electronics > Beginners |
| On a 250V capacitor or a CRT, is the negative terminal -250V? |
| (1/7) > >> |
| Starglider:
Hi, I'm doing a little explanation of how capacitors work, leading to how CRTs are basically big capacitors. Is it correct that a 250V capacitor or a CRT stores (for the sake of argument): • +250V on the positive plate/terminal • -250V on the negative Or is it 0V on the negative? What I'm trying to do is explain why you short circuit a cap or a CRT to discharge it. Mathematically: +250 -250 = 0 But if it isn't -250V on the negative then that explanation doesn't work [emoji848] Thanks! Sent from my SM-N950U1 using Tapatalk |
| Brumby:
You're understanding and associated maths needs some work - but I won't get too deep into that right now. The very first thing you need to understand is that there is no such thing as absolute voltage. Voltage is RELATIVE. What this means is that you must select a reference point before you can measure any voltage. Typically, this reference point is where you would - in practice - connect the black lead of your DMM. In the case of your capacitor, this is most commonly the "negative" terminal. When you then place the red lead to the same point, the reading will (obviously) be 0V. This is not silliness! This is the definition of the reference point. Leaving the black lead where it is, if we now move the red lead to the "positive" of your charged capacitor, it will now read 250V. That is to say, the potential on the "positive" connection of the capacitor is a positive 250V with respect to the negative terminal. Here is where it is appropriate to clarify what the +ve and -ve ends of any polarised capacitor really mean. Very simply - the positive terminal must have a voltage which is positive with respect to the negative terminal and the negative terminal must have a voltage which is negative with respect to the positive terminal. You can never say there is +250V on one terminal and -250V on the other terminal - as the difference between the two is now a mathematical 500V. This is just plain wrong. |
| Starglider:
Well, that's why I'm here. That makes a lot of sense, thank you. So it would be correct instead to say this?: "The negative plate stores -250V relative to the positive, and the positive plate stores +250V relative to the negative." If so, can you put into terms a ten year old could understand, why shorting the cap leads to it holding a charge of 0V instead of 250V? I liked the equation idea that connecting -250 +250 leaves 0. But apparently that's not right. Sent from my SM-N950U1 using Tapatalk |
| Brumby:
In the case where you have a circuit with a number of components in it, you can easily have a capacitor (or any other component for that matter) which is not connected directly to the 0V connection - as specified in a schematic, for example. It is quite possible that you could have the negative terminal of your 250V capacitor sitting at a voltage of +50V with respect to the circuit's 0V reference. With the capacitor still charged to 250V, if you measure the voltage on its positive terminal with respect to the circuit's 0V reference, you will get a reading of 300V. Similarly, you could have another circuit where the negative terminal of your 250V capacitor sitting at a voltage of -50V with respect to the circuit's 0V reference. With the capacitor still charged to 250V, if you measure the voltage on its positive terminal with respect to the circuit's 0V reference, you will get a reading of 200V. In all cases I have given, the capacitor has the same 250 volts and contains the same amount of energy. The only thing that has changed is the reference point. |
| Starglider:
Well, resetting the conversation slightly then, I simply want to explain why short circuiting a capacitor discharges it. Can you explain it better than my equation idea please? Imagine you're explaining it to a 10 year old child you know, as that's who I'm explaining it to. |
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