A couple of questions about RC and RL Circuits

[Vishera]

Question 1: why is the voltage of a capacitor as t goes to infinity equal to the voltage source? This question has been answered. The steady state voltage is not simply equal to the voltage source.

Question 2:

Just write the differential equations for voltage and current in the two situations.

V = L di/dt

I = C dv/dt

And remember that when you differentiate a sine wave, you get the cosine function. So current lags the voltage in an inductor, and voltage lags the current in a capacitor. Make sense?

So that does that mean voltage and current are always 90 degrees out of phase? I'm confused because I've done labs with RC series circuits where the voltage is only 20 degrees behind the current waveform for example. This question has been answered. They are only 90 degrees out of phase for purely capacitve or purely inductive circuits.

Question 3: According to this website, increasing the angular frequency causes the phase angle difference to increase in RC Parallel and RL Series circuits but it causes the phase angle difference to decrease in RC Series and RL Parallel circuits. Why is this? Why is the phase angle based on frequency?

Question 4: According to the this PDF, for this circuit:

http://i.imgur.com/dqjdjaf.png

the following equations apply:

http://i.imgur.com/8GUZF9D.png

Also, for this circuit:

http://i.imgur.com/dcrG29t.png

the following equations apply:

http://i.imgur.com/2WYQDsH.png

How would I derive those equations? The only equation I was given was this one:

$$v(t)=\begin{cases} v(0), & t<0 \\ v(\infty )+(v(0)-v(\infty )){ e }^{ -t/\tau }, & t>0 \end{cases}$$

and I wondering if it's possible to derive those equations from the above equation I was given.

Question 5: Is the below correct?

When a capacitor is charging, the capacitor’s voltage is at 63% and increasing when t=τ
When a capacitor is charging, the resistor’s voltage is at 37% and decreasing when t=τ

When a capacitor is discharging, the capacitor’s voltage is at 63% and decreasing when t=τ
When a capacitor is discharging, the resistor’s voltage is at 37% and increasing when t=τ

 

[psparky]

Only when the ciruit is purely inductive or purely capacitive will the current be out of phase by 90 degrees with the voltage.

Most real world situations like motors for example, there is a lot of real resitance along with the capacitance and inductance. With most of the load being resistive, the x-axis will dominate your power triangle giving you angle that are more like 20 degrees.

I just mentioned a few things about these items a couple days ago here:

https://www.physicsforums.com/showthread.php?t=746613

The transient (time) part of these caps and inductors are typically only described in DC situations. I leave that part of explanation to someone else.

And to answer the qeustion about why the capacitor meets the source voltage, that's because current no longer flows. You volt meter has no choice but to read the source since there is no voltage drop across other devices in that branch. It may not be the source voltage, more than likely a parallel branch. So it's almost similar to sticking your volt meter in a receptacle and reading a voltage.

 

[Vishera]

And to answer the qeustion about why the capacitor meets the source voltage, that's because current no longer flows. You volt meter has no choice but to read the source since there is no voltage drop across other devices in that branch. It may not be the source voltage, more than likely a parallel branch. So it's almost similar to sticking your volt meter in a receptacle and reading a voltage.

So the capacitor will only read the same value as the voltage source just as long as the capacitor is in parallel to the voltage source? What if the capacitor is in series with the voltage source? Will it still read the same as the voltage source? Or will the capacitor's voltage be equation to ${ V }_{ source }-{ V }_{ resistor }$?

 

[psparky]

So the capacitor will only read the same value as the voltage source just as long as the capacitor is in parallel to the voltage source? What if the capacitor is in series with the voltage source? Will it still read the same as the voltage source? Or will the capacitor's voltage be equation to ${ V }_{ source }-{ V }_{ resistor }$?

Basically, just follow KVL and your rules in regards to parallel devices.

If circuit only contains DC voltage source and capaictor, then capacitor voltage equals source voltage in steady state.

If I have a DC voltage source in series with a cap and a resistor, and the cap is fully charged, the voltage across the cap will equal the source because there is no current flow, therefore no voltage drop across resistor.

Lets say I have a DC voltage source in series with two resistors. R1 and R2. Now let's put that capacitor in parallel with R2. Your capacitor will NOT be equal to the voltage source in steady state. However, it will be equal to the voltage across R2. (parallel).

Picture the same scenario, but add an R3 in series with the capacitor. The cap voltage will still equal
R2 voltage...and R3 will equal zero voltage in steady state again because there is no current thru resistor R3, therefore no voltage drop.

Clear?

 

[jim hardy]

remember, phase refers to sine waves

which is what we most often encounter

but they're almost a mathematical oddity, in that differentiating them does not change their shape.

 

[psparky]

remember, phase refers to sine waves

which is what we most often encounter

but they're almost a mathematical oddity, in that differentiating them does not change their shape.

That is an interesting point. But, oh that phase shift.

When you younger students get to "bode plots" you will also see the effect of the phase change at the "break" frequencies. Bode plots refer to how a circuit will behave (gain) at different frequencies, say 0 to 20K hertz for random example. In a typical RC filter, when omega equals 1/RC, the response of a filter will change dramatically. At the break frequency in a low pass filter, that transfer function becomes 1/(1+J). There's that darn J again. In this case, you will have a 45 degree phase change. And yes, that bode plot is directly related to the speed of your RC circuit as well.

How all this stuff works is just a big puzzle. There's usually at least 5 ways to approach any problem. When you start approaching problems from all kinds of angles, things start to become more and more clear.

 

[Vishera]

Basically, just follow KVL and your rules in regards to parallel devices.

If circuit only contains DC voltage source and capaictor, then capacitor voltage equals source voltage in steady state.

If I have a DC voltage source in series with a cap and a resistor, and the cap is fully charged, the voltage across the cap will equal the source because there is no current flow, therefore no voltage drop across resistor.

Ah, that makes sense. I was confused why there is no voltage drop across the resistor.

Lets say I have a DC voltage source in series with two resistors. R1 and R2. Now let's put that capacitor in parallel with R2. Your capacitor will NOT be equal to the voltage source in steady state. However, it will be equal to the voltage across R2. (parallel).

And what will the voltage across R2 be equal to? Wouldn't it be equal to the voltage source since there is no voltage drop across R1?

So V_R2 and the capacitor's voltage would equal 10V? No, that doesn't seem right.

Picture the same scenario, but add an R3 in series with the capacitor. The cap voltage will still equal
R2 voltage...and R3 will equal zero voltage in steady state again because there is no current thru resistor R3, therefore no voltage drop.

Clear?

So R1 and R3 have 0 voltage and V_R2=V_capacitor=10V?

 

[psparky]

IN your first picture the voltage across R1 is 5 volts, R2 is 5 volts...and since cap is in parrallel, cap voltage is 5 volts. In this circuit, current flows freely thru R1 and R2...no current thru cap, just voltage.

In your 2nd picture, you did not draw what I described. R3 needs to be in same branch with cap...R3 in series with capacitor. Just flip it over to the right side of R2.

I would suggest mastering this stuff first before you move on to transient equations. Building your foundation is key before moving on. So keep asking until its crystal clear.

 

[Vishera]

IN your first picture the voltage across R1 is 5 volts, R2 is 5 volts...and since cap is in parrallel, cap voltage is 5 volts. In this circuit, current flows freely thru R1 and R2...no current thru cap, just voltage.

In your 2nd picture, you did not draw what I described. R3 needs to be in same branch with cap...R3 in series with capacitor. Just flip it over to the right side of R2.

I would suggest mastering this stuff first before you move on to transient equations. Building your foundation is key before moving on. So keep asking until its crystal clear.

Ah, I completely get it now. So for the 2nd picture with R3 in series with the cap (after the diagram is fixed), the voltage of R2 and the cap is 5V, right?

 

[psparky]

Ah, I completely get it now. So for the 2nd picture with R3 in series with the cap (after the diagram is fixed), the voltage of R2 and the cap is 5V, right?

Yes sir.

 

[sophiecentaur]

Ah, I completely get it now. So for the 2nd picture with R3 in series with the cap (after the diagram is fixed), the voltage of R2 and the cap is 5V, right?

This doesn't look right at all or perhaps we are discussing a different diagram. You have three 100Ω resistors in series so the steady state volts across each one will eventually be 10/3V. At switch on, the voltage across the capacitor (and R2) will be zero and there will be 5V across R1 ands R3

 

[psparky]

This doesn't look right at all or perhaps we are discussing a different diagram. You have three 100Ω resistors in series so the steady state volts across each one will eventually be 10/3V. At switch on, the voltage across the capacitor (and R2) will be zero and there will be 5V across R1 ands R3

Right, different diagram. If you re-read post #8, he has the circuit drawn differently then I described.

 

[sophiecentaur]

OK, that's resolved that then.
It's a good idea to draw circuit diagrams out, explicitly, when possible.