Why should care be taken while closing a capacitive circuit?What properties other than Henry and Current ratings should be taken into account when replacing inductorsQuestion on capacitive circuit in practiceWhat prevents abrupt voltage change in a circuit with a capacitor?Modeling Capacitive Discharge Ignition (CDI) CircuitOld Inductor, replace with newer current-production?How does the magnetic field of an inductor affect the operation of a DC circuit?What should be the sign of the voltage across the inductors while doing circuit analysis?Scenarios where Inductor back EMF would result in damage of circuit or component
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Why should care be taken while closing a capacitive circuit?
What properties other than Henry and Current ratings should be taken into account when replacing inductorsQuestion on capacitive circuit in practiceWhat prevents abrupt voltage change in a circuit with a capacitor?Modeling Capacitive Discharge Ignition (CDI) CircuitOld Inductor, replace with newer current-production?How does the magnetic field of an inductor affect the operation of a DC circuit?What should be the sign of the voltage across the inductors while doing circuit analysis?Scenarios where Inductor back EMF would result in damage of circuit or component
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$begingroup$
The other day in class, our professor told us that care should be taken while closing a capacitive circuit and while opening an inductive circuit? Why is this so?
I partly understand the inductive part. Since inductor only allows current to change gradually through it, if the current flow is suddenly stopped, then the voltage through it would suddenly increase, damaging it.
I'm unable to think of something for a capacitor.
capacitor inductor electrolytic-capacitor
asked 8 hours ago
user_9user_9
111 bronze badge
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
The other day in class, our professor told us that care should be taken while closing a capacitive circuit and while opening an inductive circuit? Why is this so?
I partly understand the inductive part. Since inductor only allows current to change gradually through it, if the current flow is suddenly stopped, then the voltage through it would suddenly increase, damaging it.
I'm unable to think of something for a capacitor.
capacitor inductor electrolytic-capacitor
asked 8 hours ago
user_9user_9
111 bronze badge
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Closing the circuit on a charged capacitor could, depending on the discharge path circuitry, result in a large discharge current. In particular, you do not want to be the discharge path!
$endgroup$
– Ed V
8 hours ago
$begingroup$
"Inductor only allows current to change gradually through it". Think about this: what do capacitors prefer to change gradually?
$endgroup$
– glen_geek
8 hours ago
$begingroup$
@glen_geek Voltage right? If voltage changes quickly, the current would suddenly increase. What would happen to the capacitor then?
$endgroup$
– user_9
7 hours ago
$begingroup$
Yes, high current is the result of fast-changing voltage. Every capacitor has some hopefully small internal resistance (ESR)...high currents must flow through this resistance generating heat. Not a desirable thing.
$endgroup$
– glen_geek
5 hours ago
add a comment |
$begingroup$
The other day in class, our professor told us that care should be taken while closing a capacitive circuit and while opening an inductive circuit? Why is this so?
I partly understand the inductive part. Since inductor only allows current to change gradually through it, if the current flow is suddenly stopped, then the voltage through it would suddenly increase, damaging it.
I'm unable to think of something for a capacitor.
capacitor inductor electrolytic-capacitor
asked 8 hours ago
user_9user_9
111 bronze badge
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The other day in class, our professor told us that care should be taken while closing a capacitive circuit and while opening an inductive circuit? Why is this so?
I partly understand the inductive part. Since inductor only allows current to change gradually through it, if the current flow is suddenly stopped, then the voltage through it would suddenly increase, damaging it.
I'm unable to think of something for a capacitor.
capacitor inductor electrolytic-capacitor
capacitor inductor electrolytic-capacitor
asked 8 hours ago
user_9user_9
111 bronze badge
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
user_9user_9
111 bronze badge
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
user_9user_9
111 bronze badge
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
user_9user_9
111 bronze badge
asked 8 hours ago
user_9user_9
111 bronze badge
111 bronze badge
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user_9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Closing the circuit on a charged capacitor could, depending on the discharge path circuitry, result in a large discharge current. In particular, you do not want to be the discharge path!
$endgroup$
– Ed V
8 hours ago
$begingroup$
"Inductor only allows current to change gradually through it". Think about this: what do capacitors prefer to change gradually?
$endgroup$
– glen_geek
8 hours ago
$begingroup$
@glen_geek Voltage right? If voltage changes quickly, the current would suddenly increase. What would happen to the capacitor then?
$endgroup$
– user_9
7 hours ago
$begingroup$
Yes, high current is the result of fast-changing voltage. Every capacitor has some hopefully small internal resistance (ESR)...high currents must flow through this resistance generating heat. Not a desirable thing.
$endgroup$
– glen_geek
5 hours ago
add a comment |
$begingroup$
Closing the circuit on a charged capacitor could, depending on the discharge path circuitry, result in a large discharge current. In particular, you do not want to be the discharge path!
$endgroup$
– Ed V
8 hours ago
$begingroup$
"Inductor only allows current to change gradually through it". Think about this: what do capacitors prefer to change gradually?
$endgroup$
– glen_geek
8 hours ago
$begingroup$
@glen_geek Voltage right? If voltage changes quickly, the current would suddenly increase. What would happen to the capacitor then?
$endgroup$
– user_9
7 hours ago
$begingroup$
Yes, high current is the result of fast-changing voltage. Every capacitor has some hopefully small internal resistance (ESR)...high currents must flow through this resistance generating heat. Not a desirable thing.
$endgroup$
– glen_geek
5 hours ago
$begingroup$
Closing the circuit on a charged capacitor could, depending on the discharge path circuitry, result in a large discharge current. In particular, you do not want to be the discharge path!
$endgroup$
– Ed V
8 hours ago
$begingroup$
Closing the circuit on a charged capacitor could, depending on the discharge path circuitry, result in a large discharge current. In particular, you do not want to be the discharge path!
$endgroup$
– Ed V
8 hours ago
$begingroup$
"Inductor only allows current to change gradually through it". Think about this: what do capacitors prefer to change gradually?
$endgroup$
– glen_geek
8 hours ago
$begingroup$
"Inductor only allows current to change gradually through it". Think about this: what do capacitors prefer to change gradually?
$endgroup$
– glen_geek
8 hours ago
$begingroup$
@glen_geek Voltage right? If voltage changes quickly, the current would suddenly increase. What would happen to the capacitor then?
$endgroup$
– user_9
7 hours ago
$begingroup$
@glen_geek Voltage right? If voltage changes quickly, the current would suddenly increase. What would happen to the capacitor then?
$endgroup$
– user_9
7 hours ago
$begingroup$
Yes, high current is the result of fast-changing voltage. Every capacitor has some hopefully small internal resistance (ESR)...high currents must flow through this resistance generating heat. Not a desirable thing.
$endgroup$
– glen_geek
5 hours ago
$begingroup$
Yes, high current is the result of fast-changing voltage. Every capacitor has some hopefully small internal resistance (ESR)...high currents must flow through this resistance generating heat. Not a desirable thing.
$endgroup$
– glen_geek
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In theory when an empty capacitor is connected to a ideal voltage source, infinite current flows until capacitor is charged to the supplied voltage. Basically same thing when disconnecting a powered inductor, in theory the voltage goes to infinity. Many devices like computer power supplies, LED lamps and phone chargers have inrush current limiting because of capacitance.
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
$endgroup$
$begingroup$
But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit.
$endgroup$
– Uwe
8 hours ago
add a comment |
$begingroup$
Vc = Ic *ESR + C * dVc/dt. Where Ic=Vcc*ESR. So if Vcc =12V, and ultra low ESR was 3 milliohm that in theory = 1kA If Vcc is closer to ideal than cap. which creates 3kW pulse. That might be extreme but illustrates the math .. otherwise the source Vcc would drop as C cgarges up.
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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votes
$begingroup$
In theory when an empty capacitor is connected to a ideal voltage source, infinite current flows until capacitor is charged to the supplied voltage. Basically same thing when disconnecting a powered inductor, in theory the voltage goes to infinity. Many devices like computer power supplies, LED lamps and phone chargers have inrush current limiting because of capacitance.
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
$endgroup$
$begingroup$
But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit.
$endgroup$
– Uwe
8 hours ago
add a comment |
$begingroup$
In theory when an empty capacitor is connected to a ideal voltage source, infinite current flows until capacitor is charged to the supplied voltage. Basically same thing when disconnecting a powered inductor, in theory the voltage goes to infinity. Many devices like computer power supplies, LED lamps and phone chargers have inrush current limiting because of capacitance.
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
$endgroup$
$begingroup$
But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit.
$endgroup$
– Uwe
8 hours ago
add a comment |
$begingroup$
In theory when an empty capacitor is connected to a ideal voltage source, infinite current flows until capacitor is charged to the supplied voltage. Basically same thing when disconnecting a powered inductor, in theory the voltage goes to infinity. Many devices like computer power supplies, LED lamps and phone chargers have inrush current limiting because of capacitance.
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
$endgroup$
In theory when an empty capacitor is connected to a ideal voltage source, infinite current flows until capacitor is charged to the supplied voltage. Basically same thing when disconnecting a powered inductor, in theory the voltage goes to infinity. Many devices like computer power supplies, LED lamps and phone chargers have inrush current limiting because of capacitance.
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
answered 8 hours ago
JustmeJustme
6,3412 gold badges6 silver badges17 bronze badges
6,3412 gold badges6 silver badges17 bronze badges
$begingroup$
But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit.
$endgroup$
– Uwe
8 hours ago
add a comment |
$begingroup$
But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit.
$endgroup$
– Uwe
8 hours ago
$begingroup$
But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit.
$endgroup$
– Uwe
8 hours ago
$begingroup$
But also when the current flowing into the capacitor is limited by a resistor, the maximum of current is reached instantly after closing the circuit.
$endgroup$
– Uwe
8 hours ago
add a comment |
$begingroup$
Vc = Ic *ESR + C * dVc/dt. Where Ic=Vcc*ESR. So if Vcc =12V, and ultra low ESR was 3 milliohm that in theory = 1kA If Vcc is closer to ideal than cap. which creates 3kW pulse. That might be extreme but illustrates the math .. otherwise the source Vcc would drop as C cgarges up.
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
$endgroup$
add a comment |
$begingroup$
Vc = Ic *ESR + C * dVc/dt. Where Ic=Vcc*ESR. So if Vcc =12V, and ultra low ESR was 3 milliohm that in theory = 1kA If Vcc is closer to ideal than cap. which creates 3kW pulse. That might be extreme but illustrates the math .. otherwise the source Vcc would drop as C cgarges up.
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
$endgroup$
add a comment |
$begingroup$
Vc = Ic *ESR + C * dVc/dt. Where Ic=Vcc*ESR. So if Vcc =12V, and ultra low ESR was 3 milliohm that in theory = 1kA If Vcc is closer to ideal than cap. which creates 3kW pulse. That might be extreme but illustrates the math .. otherwise the source Vcc would drop as C cgarges up.
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
$endgroup$
Vc = Ic *ESR + C * dVc/dt. Where Ic=Vcc*ESR. So if Vcc =12V, and ultra low ESR was 3 milliohm that in theory = 1kA If Vcc is closer to ideal than cap. which creates 3kW pulse. That might be extreme but illustrates the math .. otherwise the source Vcc would drop as C cgarges up.
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
answered 20 mins ago
Sunnyskyguy EE75Sunnyskyguy EE75
80.1k2 gold badges30 silver badges116 bronze badges
80.1k2 gold badges30 silver badges116 bronze badges
add a comment |
add a comment |
user_9 is a new contributor. Be nice, and check out our Code of Conduct.
user_9 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Closing the circuit on a charged capacitor could, depending on the discharge path circuitry, result in a large discharge current. In particular, you do not want to be the discharge path!
$endgroup$
– Ed V
8 hours ago
$begingroup$
"Inductor only allows current to change gradually through it". Think about this: what do capacitors prefer to change gradually?
$endgroup$
– glen_geek
8 hours ago
$begingroup$
@glen_geek Voltage right? If voltage changes quickly, the current would suddenly increase. What would happen to the capacitor then?
$endgroup$
– user_9
7 hours ago
$begingroup$
Yes, high current is the result of fast-changing voltage. Every capacitor has some hopefully small internal resistance (ESR)...high currents must flow through this resistance generating heat. Not a desirable thing.
$endgroup$
– glen_geek
5 hours ago