Just apply simple KVL, U have 3 voltages , two of them are known and one is unknown and the sum of the voltages would equal zero.
I think It's very easy, U only need to take care of applying KVL and the sign of each voltage!!!

There is a real issue with how to interpret the information given in the problem. Most specifically, what that meter is showing. If it is showing RMS voltage, then the problem is inconsistently overdefined. If it is showing instantaneous voltage, then you don't know what the instantaneous voltage output from the AC source is because you don't know the time (but you do have enough information to find it. If, as it appears crutschow might be interpreting it, that is not a meter but a second DC source, then the resistor R1 is effectively removed and the voltage across the cap is just the difference between the two sources. Given the low impedance of C1 (about -j16Ω), the impedance of the two sources may be a factor, but unless they are more than an ohm or two combined, the effect will be minimal.

 

Could you please sketch a schematic, because I can't picture what you have in mind. Now you are talking about two sources. Where is this second source coming from? Are you thinking that the meter across R1 is a voltage source?

Okay, we've come to the source of the (my) confusion.
I assumed it was a source as it didn't make sense to me that it was a meter, since the signal is AC and the box has a polarity shown at its terminals.
But It could be a meter. Don't see what it would be measuring though as it appears to be a DC meter.

 

Okay, we've come to the source of the (my) confusion.
I assumed it was a source as it didn't make sense to me that it was a meter, since the signal is AC and the box has a polarity shown at its terminals.
But It could be a meter. Don't see what it would be measuring though as it appears to be a DC meter.

I think it is a DC meter and that the intent of the question is to find the instantaneous voltage across the capacitor at the moment in time that the instantaneous voltage across the resistor is as shown. I haven't cranked the math to see if there is a single answer to that question, assuming that that is the intended interpretation. It would sure be nice if the OP would clarify things, but I suspect we have another one-hit wonder.

 

Thanks!

 

Instead of saying thanks, how about clarifying the problem for us. Namely, what information is that meter giving you and what are you being asked to calculate? Is it the instantaneous voltage across the resistor at the moment in time that you are being asked for the instantaneous voltage across the capacitor?

 

Instead of saying thanks, how about clarifying the problem for us. Namely, what information is that meter giving you and what are you being asked to calculate? Is it the instantaneous voltage across the resistor at the moment in time that you are being asked for the instantaneous voltage across the capacitor?

Actually I am not clear about the problem. I just calculated it as 10-7.072 =2.298. I also tried for finding out equivalent impedance then total current and then voltage as in 1st reply. The answer does not match. Our method is right. But answer is 111mV. The problem/answer given may be wrong. I am not sure.

 

As I stated in post #8, you have too many numbers and the problem is incorrectly stated.

Remove the meter that reads 7.072V and then you can solve the problem.

 

As I stated in post #8, you have too many numbers and the problem is incorrectly stated.

Remove the meter that reads 7.072V and then you can solve the problem.

What if, instead of a meter across R1, there was a clock saying that the time was 16.1 us? Would you still say that there are two many numbers?

 

Then I would have to say there are too few numbers. We don't know the phase angle of the power supply.

 

Then I would have to say there are too few numbers. We don't know the phase angle of the power supply.

If a specific phase angle is not given, then the general assumption is that it is zero degrees. But even that isn't needed in this case since if you use that meter as a means of determining the time, then it takes into account the phase angle of the power supply, doesn't it?

 

Actually I am not clear about the problem. I just calculated it as 10-7.072 =2.298. I also tried for finding out equivalent impedance then total current and then voltage as in 1st reply. The answer does not match. Our method is right. But answer is 111mV. The problem/answer given may be wrong. I am not sure.

If you ignore the meter and just find the RMS voltage across the capacitor, what do you get? I get something between 150 mVrms and 175 mVrms.

If you assume that the question is asking for the instantaneous voltage across the capacitor at the time when the instantaneous voltage across the resistor is as indicated by the meter, what do you get? I get ±(something a bit under 200 mV).

Neither match the 111 mV answer given.

 

I got similar values you mentioned. n not 111mV.

 

Reactance: XC = 1/jwC
Resistance: R
Impedance: Z = XC + R

Momentary voltage from source: VR = VS * (R/Z) => VS = VR * (Z/R)

Momentary voltage across capacitor: VC = VS * (XC/Z) [not considering -90 degree voltage displacement]

Think this is correct, but it may be wrong.

So the right answer should be 111 mV?

Reactance: XC = 1/jwC = 1/(j*2*pi*10^4*10^(-6)) = 15.92 Ohms
Resistance: R = 1000 Ohms
Impedance: Z = XC + R = 1015.92 Ohms

Momentary voltage from source: VR = VS * (R/Z) => VS = VR * (Z/R) = 7.072 * (1000/1015.92) = 7.185 V

Momentary voltage across capacitor: VC = VS * (XC/Z) = 7.185 * (15.92/1015.92) = 0.113 V [not considering -90 degree voltage displacement]

Pretty close if you ask me.

 

So the right answer should be 111 mV?

Reactance: XC = 1/jwC = 1/(j*2*pi*10^4*10^(-6)) = 15.92 Ohms
Resistance: R = 1000 Ohms
Impedance: Z = XC + R = 1015.92 Ohms

Momentary voltage from source: VR = VS * (R/Z) => VS = VR * (Z/R) = 7.072 * (1000/1015.92) = 7.185 V

Momentary voltage across capacitor: VC = VS * (XC/Z) = 7.185 * (15.92/1015.92) = 0.113 V [not considering -90 degree voltage displacement]

Pretty close if you ask me.

You've got a couple of problems here. First impedance is not the arithmetic sum of reactance and resistance. The magnitude of the impedance is the Pythagorean sum of the two, namely

\(
\left| Z \right| \; = \; \sqrt{R^2 \; + \; X_C^2}
\)

In this case, because R >> Xc, the two come out pretty close.

More importantly, however, is that you got an answer close to theirs by not considering the 90 degree phase shift in the voltage. But there IS a phase shift that can't be "not considered".

 

You've got a couple of problems here. First impedance is not the arithmetic sum of reactance and resistance. The magnitude of the impedance is the Pythagorean sum of the two, namely

\(
\left| Z \right| \; = \; \sqrt{R^2 \; + \; X_C^2}
\)

In this case, because R >> Xc, the two come out pretty close.

More importantly, however, is that you got an answer close to theirs by not considering the 90 degree phase shift in the voltage. But there IS a phase shift that can't be "not considered".

Gotcha. Just thought I had to include this as I remembered getting something close to the answer. I have no problem being schooled, as long as it's done in a respectful manner. Thank you.

Edit: The reason I omitted the phase shift is because it looked like a "starter"-task. Seeing no one has solved it yet, it may not be much of a "starter"-task after all

 

Thank you all!

 

You're more than welcome.

If you get the "official" solution, please share so that, if nothing else, we can debate how the question should have been presented.