Understanding Capacitive Circuits: The Relationship Between Current and Voltage

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SUMMARY

The discussion clarifies the concept of "current leads the voltage by 90 degrees" in purely capacitive circuits. It emphasizes that in capacitors, the current must flow to charge the capacitor before a voltage can develop across its terminals, establishing a causal relationship. The differential equations governing capacitors and inductors are referenced to explain the phase relationship between current and voltage. This terminology, while potentially confusing, accurately describes the behavior of AC circuits involving capacitors.

PREREQUISITES
  • Understanding of AC circuit fundamentals
  • Familiarity with differential equations in electrical engineering
  • Knowledge of capacitor behavior and charging dynamics
  • Basic concepts of inductors and their voltage-current relationship
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  • Study the differential equations for capacitors and inductors in detail
  • Learn about the phase relationships in AC circuits
  • Explore the implications of current and voltage relationships in reactive components
  • Investigate practical applications of capacitive circuits in power supplies
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Electrical engineering students, educators teaching AC circuit theory, and professionals working with capacitive components in circuit design.

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Perhaps some one can explain to me what is meant by "current leads the voltage by 90 degrees"...let this inquiry be in regards to a purely capacitive circuit.

What seems to baffle me is why this terminology is used. Its not like the current just decided to occur before a potenial difference.
 
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XPTPCREWX said:
Perhaps some one can explain to me what is meant by "current leads the voltage by 90 degrees"...let this inquiry be in regards to a purely capacitive circuit.

What seems to baffle me is why this terminology is used. Its not like the current just decided to occur before a potenial difference.

It's easiest to get comfortable with this concept based on the differential equations for the currents and voltages for inductors and capacitors.

For Capacitors, the driving function is the current, which charges up the capacitors:

v(t) = \frac{1}{C}\int{i(t) dt}

The voltage across a capacitor cannot change instantaneously, because that would require an infinite current.

Similarly for Inductors, the driving function is the voltage, which ramps up the current:

i(t) = \frac{1}{L}\int{v(t) dt}

The current in the inductor cannot change instantaneously, bacause that would require an infinite voltage. The current can change very quickly, however, like in a flyback power supply circuit that is pumping a high voltage...

So when we say that the current lags the applied voltage in inductors, or that the voltage lags the applied current in capacitors, we are just saying in words what you get from the differential equations that describe the physics of those components.

And when we say "the current leads the voltage...", we are not describing something non-causal. It's just a bit of a confusing way to state the relationship the other way around, and for a continuous waveform.

Hope that helps.
 
Last edited:
Really helps...Thanks.
 
XPTPCREWX said:
Perhaps some one can explain to me what is meant by "current leads the voltage by 90 degrees"...let this inquiry be in regards to a purely capacitive circuit.

What seems to baffle me is why this terminology is used. Its not like the current just decided to occur before a potenial difference.

To make it very simple. Before the cap can have any voltage across it's terminals, you need to charge it up. So current has to be pumped into the cap before the cap can charge up to have voltage. Current always have to come first so it lead the voltage! This is the ABC way to explain it!:smile:

Berkeman has given you the detail of the rest.
 
yungman said:
To make it very simple. Before the cap can have any voltage across it's terminals, you need to charge it up. So current has to be pumped into the cap before the cap can charge up to have voltage. Current always have to come first so it lead the voltage! This is the ABC way to explain it!:smile:

Berkeman has given you the detail of the rest.

yungman,

Actually it isn't that simple. there is obviously a potential difference between the applied voltage and the capacitors terminal voltage to create that current flow in the first place.

Berkeman has explained it to me the best so far.
 
XPTPCREWX said:
yungman,

Actually it isn't that simple. there is obviously a potential difference between the applied voltage and the capacitors terminal voltage to create that current flow in the first place.

Berkeman has explained it to me the best so far.


I am glad you are happy with berkeman's answer. This is a very simple question that you find in the first few lessons of AC circuits, so I want to give a very simple answer just in case. I just add this in because there are people that don't do calculus! I was one of those long time ago!
 
Most likely this can only be answered by an "old timer". I am making measurements on an uA709 op amp (metal can). I would like to calculate the frequency rolloff curves (I can measure them). I assume the compensation is via the miller effect. To do the calculations I would need to know the gain of the transistors and the effective resistance seen at the compensation terminals, not including the values I put there. Anyone know those values?

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