Electronics - RC Circuits - Discussion

3. 

What is the voltage drop across R1 in the given circuit?

[A]. 10 V
[B]. 4.80 V
[C]. 4.00 V
[D]. 5.80 V

Answer: Option D

Explanation:

No answer description available for this question.

Raja said: (Sep 4, 2011)  
R2 ll 250i +240i + 180.

Here voltage divides.

Ramkumar said: (Oct 21, 2011)  
Please explain briefly? I don't understand.

Ajayvignesh said: (Nov 11, 2011)  
Raja you are wrong. R2 is not parallel to summation of all three. its parallel to only XC2. Hence parallel reistance is 142.8Ohm. Hence drop across 142.8 (equivalanet resistance of 333Ohm and XC2 250Ohm) is

Ajay said: (Dec 29, 2011)  
Its very easy raja your absolutly correct. This is very basic.

Shahid said: (Sep 26, 2012)  
Hey friends let me explain.
For rc SERIES circuit Z=Sqrt of R^2+xC^2
R2 is parallel to 250 so it will be 142.79
It is in series with 180 so total res Rt will be 322.79
Now use formula=z=sqrt(322.79^2+240^2)=402.23
Now we have I=V/Z=0.0248
Now vtg drop across r1=0.024*(180/180+240)=5.78v

Farhan said: (Jul 7, 2013)  
@Shahid.

You are doing totally wrong, Now you see your step where you find the voltage drop across R1 you use current divider formula that's wrong applied. You use voltage divider instead of current divider.

Vdrop_R10= (R10) *VSupply /Total Resistance of the circuit.

Jay said: (Feb 13, 2014)  
Here,
I = V/Req.
Req. = R1+XC1+R2 parallel with XC2.

Voltage across R1 is = I*R1.

Scott said: (May 29, 2014)  
I = V/Req.
Zt = R1+XC1+R2 parallel with XC2.
Zt = 500 ohm @ -53.1 degrees.
It = 10v/500 ohm @ -53.1 degrees = 20 mA @ 53.1 degrees.
Vr1 = 180 ohms @ 0 degrees * 20 mA @ 53.1 degrees = 3.6 V @ 53.1 degrees.

Archana Sasikumar said: (Jun 25, 2014)  
R2||XC2 = 142.79 OHM.
XC1+(R2||XC2) = 382.79 OHM.
V1 = 10*(180/180+382.79)V.

Balram said: (Aug 28, 2014)  
Please provide calculation step by step with logic and formula applied.

Sangamesh said: (May 31, 2015)  
Since the parallel combination of xc2 and R2 = 142.79.

Therefore z = square root(180^2+240^2+142.79^2) = 332.25.

Hence Vr1 = 10*180/332.25 = 5.8v.

Alex said: (Aug 28, 2015)  
No. Just simply, no.

You guys are all doing this wrong and the answers are wrong.

You can't simply combine the parallel R2||Xc2 in the traditional method. You have to remember that capacitive reactance is at an angle of -90°.

When you combine that parallel branch, you have to convert back and forth between polar and rectangular forms as appropriate (addition in rectangular, multiplication/division in polar).

So let me walk you through this:

Req=R2||X2.
Req=333||-j250.

Req=1/((1/333)+(1/250<-90°)) <--in polar for division.
Req=1/(0.003+j 0.004) <--in rectangular for addition.

Req=1/(0.005<53°) <--Convert to a vector/rectangular.
Req=200<-53° <-- That is the equivalent resistance.

Now, convert that back to rectangular and add R1 and Xc1:

200<-53° = 120-j159.73.

Zt = (120-j159.73)+(180-j240).

Zt = 300-j400 <-- (rounded to -j400).

Zt = 500<-53° <-- (polar representation).

Now:

It = Vt/Zt=10V/500<-53°) = 0.02 A < 53°.

VR1 = It*R1 = 0.02 A*180Ω = 3.6 V.

Abhiee said: (Oct 19, 2015)  
Let me tell in more simpler way.

R1 and XC1 are series to source. So z1 = sqrt (R1^2+XC1^2).

R2 and XC2 are parallel to source. So z2 = 1/sqrt ((1/R2)^2+(1/XC2)^2).

Then z = z1+z2. Voltage drop = R1/z*vs.

Privusr said: (Apr 3, 2016)  
@Alex, your answer is 3.6 V. They give 5.80 V.

Which one is correct? Anybody tell me.

Divyanshu Shekhar said: (Nov 22, 2016)  
I think here XL1 = 240 ohm not XC 1=240 ohm.

When XL 1 = 240 ohm, then calculate:

z1 = 180 + j240
z2 = 333 * -j250/(333-j250)
= 120.03 -j159.88
= 120-j160 (approx.)
z = z1 + z2
= 300 + j80

z = sqrt[(300)2 + (80)2]
=310.48 ohm = 310.5 ohm(approx.).

i = v/z = 10/310.5 = 0.0322 A.
Voltage drop across r1 = 180 * 0.0322.
= 5.79 volts.
= 5.8 volt(approx).

Also, phase angle, tan(θ) = 80/300 = 0.26666 = 0.267(approx).
So, θ = Arctan(0.267).
= 14.95 degree.

Rose said: (Nov 25, 2016)  
I want an answer to my question.

Suppose we have a circuit where the ratio of the complex amplitude of the particular integral to the amplitude of the driving sinusoid is

VpVi = jωRC1 + jωRC.

Given that R = 0.82kΩ and C = 10.0nF we will compute the magnitude and phase of this ratio for various frequencies.

For frequency f = 170Hz what is the magnitude of VpVi?

Safna said: (Jan 4, 2018)  
Please explain.

Samar said: (Jan 28, 2019)  
Here;

1. Get the inverse (1/333 - 1/250i)^-1 to get the impedance since it is in admittance form, then it will be equal to 120.03 + 159.88i.
2. Then next step is to add that to (180 - 240i).
3. And, (120.03 + 159.88i) + (180 - 240i) = 300.09 - 80.12i.
4. Then convert that to Polar form which is equal to 310.55 <-14.95.
5. Now the total impedance is 310.55, and by voltage divider formula Vr1 = (10/310.55)(180) = 5.8 volts.

Grandaphi said: (Aug 21, 2019)  
Doesn't (1/333-1/250i)^-1 = 120.03 - 159.88i not 120 + 159.88i?

And I get 3.6V.

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