How is the voltage across the Capacitor the Voltage across the 35Ω resistor

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
70
In the professors solution to question 6 part (b) I don't understand why the voltage Vc is the same as the voltage across the 35Ω resistor, is it because the capacitor and the 35Ω resistor are parallel to each other?

I thought that if the Capacitor was at Steady State DC then it behaved like an open circuit, so if no current is flowing how can there be a voltage drop?

2009 dec EC2 q6.PNG
2009 dec EC2 q6 part solution.PNG
 

Irving

Joined Jan 30, 2016
946
The lower left picture says it all... at DC the capacitor is open circuit, and the inductor is a short circuit. therefore the 35ohm resistor is effectively in parallel with the capacitor. The voltage drop is due to the current through the 35 ohm resistor which is the voltage appearing across the capacitor terminals even though it has no action in this circuit at DC steady state..

Review your Kirchoff voltage laws... :)
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
70
The lower left picture says it all... at DC the capacitor is open circuit, and the inductor is a short circuit. therefore the 35ohm resistor is effectively in parallel with the capacitor. The voltage drop is due to the current through the 35 ohm resistor which is the voltage appearing across the capacitor terminals even though it has no action in this circuit at DC steady state..

Review your Kirchoff voltage laws... :)
Thank you so much, I will look at KVL again, I get so confused by this because I don't understand how there can be a voltage drop without current flow but of course there must be for the capacitor to have charged. I just remember my professor saying there can't be zero current if there is a voltage drop, is that true?
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
70
hi Mojo,
Look at this redraw, does it make it clear.
E
Thank you, that does really help, I understand how to answer the question now, I think what I find tough is how the capacitor is open circuit and no current can flow but there is a voltage drop across it. Would this question be answered similarly if I had to find the voltage across the inductor?
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
70
hi mojo,
Don't confuse 'voltage drop' with 'applied voltage '.
What is the source of the 'voltage applied' across the capacitor.?
E
ok, I'm not sure, but I'll have a try. The voltage applied to the capacitor is the provided by the 15mA current source which I think comes makes the circuit voltage 200mV. Does this mean that you can apply a voltage at one side of the capacitor and there can be another voltage at the other side not because it 'dropped across the capacitor' but because the current flow through the parallel 35Ω resistor caused the voltage to drop?
 

WBahn

Joined Mar 31, 2012
26,062
Thank you so much, I will look at KVL again, I get so confused by this because I don't understand how there can be a voltage drop without current flow but of course there must be for the capacitor to have charged. I just remember my professor saying there can't be zero current if there is a voltage drop, is that true?
Are you sure he wasn't making that statement in the context of a resistive element?

A voltage drop just says that two points are at different electrical potentials. There's a voltage drop of 9 V across the terminals of a 9 V battery that is sitting on the shelf with nothing connected to it. So you have a voltage drop without a current.

The relationship between voltage and current is dictated by the physics of the devices involved and is described by that device's constitutive equation.

For a resistor, that is Ohm's Law,

v(t) = i(t)·R

So the only way to have a non-zero voltage without a non-zero current is for R to be infinite.

But for a capacitor, the constitutive equation is

i(t) = C·dv(t)/dt

So i(t) = 0 any time that v(t) is not changing, but other than that v(t) can be any value, positive, negative, or zero.
 

WBahn

Joined Mar 31, 2012
26,062
ok, I'm not sure, but I'll have a try. The voltage applied to the capacitor is the provided by the 15mA current source which I think comes makes the circuit voltage 200mV. Does this mean that you can apply a voltage at one side of the capacitor and there can be another voltage at the other side not because it 'dropped across the capacitor' but because the current flow through the parallel 35Ω resistor caused the voltage to drop?
Don't make it harder than it is. The term "drop" is just an analogous use of the term from our everyday experience of something that involves two points.

If you are on a mountain and there's a cabin at one point and a shed at another point, you might say something like, "There's a 100 ft drop from the shed to the cabin," or you might say that it's a 1000 ft drop from the top of the cliff to the bottom." In neither case is it required that anything actually be dropping for the statement to be well-defined and make sense.

In a gravitational field, elevation plays the same role as voltage in a (conservative) electric field, so we can use some of the terms in one to describe things in the other.
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
70
Are you sure he wasn't making that statement in the context of a resistive element?

A voltage drop just says that two points are at different electrical potentials. There's a voltage drop of 9 V across the terminals of a 9 V battery that is sitting on the shelf with nothing connected to it. So you have a voltage drop without a current.

The relationship between voltage and current is dictated by the physics of the devices involved and is described by that devices constitutive equation.

For a resistor that is Ohm's Law,

v(t) = i(t)·R

So the only way to have a non-zero voltage without a non-zero current is for R to be infinite.

But for a capacitor, the constitutive equation is

i(t) = C·dv(t)/dt

So i(t) = 0 any time that v(t) is not changing, but other than that v(t) can be any value, positive, negative, or zero.
This made it click, thank you
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
70
Don't make it harder than it is. The term "drop" is just an analogous use of the term from our everyday experience of something that involves two points.

If you are on a mountain and there's a cabin at one point and a shed at another point, you might say something like, "There's a 100 ft drop from the shed to the cabin," or you might say that it's a 1000 ft drop from the top of the cliff to the bottom." In neither case is it required that anything actually be dropping for the statement to be well-defined and make sense.

In a gravitational field, elevation plays the same role as voltage in a (conservative) electric field, so we can use some of the terms in one to describe things in the other.
I was overthinking it, I appreciate your help, thank you
 

OBW0549

Joined Mar 2, 2015
3,506
This made it click, thank you
Another thing that may help for quickie analysis purposes is to remember that at DC (that is, at steady state, after voltages and currents have settled down to their final values), capacitors can be replaced by an open circuit (i.e., removed) and inductors can be replaced by a short (i.e., a piece of wire).

Commit these two relations to memory, right up there alongside Ohm's Law:

For a capacitor, I(c) = C * dV / dT (amps, farads, volts, & seconds) and
For an inductor, V(l) = L * dI / dT (volts, henries, amps, & seconds).

Or, verbally,

For a capacitor, the current through a capacitor is equal to the capacitance times the rate of change of the voltage across it; and for an inductor, the voltage across an inductor is equal to the inductance times the rate of change of the current flowing through it.
 

Thread Starter

Mojo Pin__

Joined Apr 13, 2019
70
But for a capacitor, the constitutive equation is

i(t) = C·dv(t)/dt

So i(t) = 0 any time that v(t) is not changing, but other than that v(t) can be any value, positive, negative, or zero.
Can I just check with you please,

i(t) = 0 when v(t) is not changing AND when v(t) is zero, right?

or is there a case where v(t) can equal zero and we get a non zero value for i(t)
 

WBahn

Joined Mar 31, 2012
26,062
Can I just check with you please,

i(t) = 0 when v(t) is not changing AND when v(t) is zero, right?

or is there a case where v(t) can equal zero and we get a non zero value for i(t)
Imagine v(t) going smoothly from some negative value to some positive value. Even as v(t) passes through 0 V, it is still changing, and hence i(t) will be non-zero (assuming C is non-zero).

In fact, if the circuit is an AC circuit being driven by a sinusoidal waveform, the current will be a maximum when the voltage is zero and, conversely, will be zero when the voltage is at a max.
 

Irving

Joined Jan 30, 2016
946
All of the above is good, but I think the OP needs to clarify terminology in his head.

With my students I ask them to use 'potential difference' (PD) when talking about voltages as its correct for any scenario, whereas 'voltage drop' (VD) is, arguably, a dynamic thing relating to the action of, typically, current in a resistor. So VD referring to a battery's output terminals is syntactically incorrect as the battery is the source of the PD. But either VD or PD ACROSS the 35 ohm resistor is clearly and unequivocally as a result of the current through it. To use the term VD when referring to the capacitor is both meaningless and confusing as it plays no part in that. It does, however, experience a PD due to the PD/VD ACROSS the resistor.

In short there can never be a VD across a capacitor, its always a PD reflecting the state of charge of the capacitor. By analogy there is never a physical current flow through a capacitor; there is a change of charge on the capacitor's plates, but electrons never go through the capacitor, so to talk about the current 'through' a capacitor is, IMHO, technically meaningless.
 

WBahn

Joined Mar 31, 2012
26,062
All of the above is good, but I think the OP needs to clarify terminology in his head.

With my students I ask them to use 'potential difference' (PD) when talking about voltages as its correct for any scenario, whereas 'voltage drop' (VD) is, arguably, a dynamic thing relating to the action of, typically, current in a resistor. So VD referring to a battery's output terminals is syntactically incorrect as the battery is the source of the PD. But either VD or PD ACROSS the 35 ohm resistor is clearly and unequivocally as a result of the current through it. To use the term VD when referring to the capacitor is both meaningless and confusing as it plays no part in that. It does, however, experience a PD due to the PD/VD ACROSS the resistor.

In short there can never be a VD across a capacitor, its always a PD reflecting the state of charge of the capacitor. By analogy there is never a physical current flow through a capacitor; there is a change of charge on the capacitor's plates, but electrons never go through the capacitor, so to talk about the current 'through' a capacitor is, IMHO, technically meaningless.
Even if I were to stipulate to everything you've just claimed, the OP also needs to be able to work in the real world using terms as they are used in the real world. In the real world, "voltage drop" and "potential difference" are almost always used synonymously.

As for the current through a capacitor, yes, electrons don't flow through the capacitor's dielectric, but the displacement current does.
 
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