Electrical Engineering - RC Circuits - Discussion
Discussion Forum : RC Circuits - General Questions (Q.No. 19)
19.
In a series RC circuit, when the frequency and the resistance are halved, the impedance
Discussion:
6 comments Page 1 of 1.
Joesmarti said:
3 years ago
Z = R - jX.
R' = 0.5 R.
X' = 1 / 2* π*C*(0.5)f = 2X.
Z' = 0.5R - j2X.
Assuming R = 1 and X = 1.
|Z| = √(1^2+1^2) = 1.414.
|Z'| = √(0.5^2+2^2) = 2.0615.
R' = 0.5 R.
X' = 1 / 2* π*C*(0.5)f = 2X.
Z' = 0.5R - j2X.
Assuming R = 1 and X = 1.
|Z| = √(1^2+1^2) = 1.414.
|Z'| = √(0.5^2+2^2) = 2.0615.
Noor ul islam said:
7 years ago
z =√((2R)^2+(2/wC)^2) because resistance is double and frequency is half.
z = 2√((2R)^2+(2/wC)^2).
So the impedence is double.
z = 2√((2R)^2+(2/wC)^2).
So the impedence is double.
Manish said:
8 years ago
There is no explanation about capacitance so without value of capacitance we can't say anything.
Dhaval said:
8 years ago
Here, Z=R+Xc.
Now resistance and frequency is halved so R become R/2 and Xc=1/2π(f/2)c => 2Xc=1/2πfc.
Z'=R/2+ 2Xc so impedance neither double nor halved. So, It cannot determine without values.
Now resistance and frequency is halved so R become R/2 and Xc=1/2π(f/2)c => 2Xc=1/2πfc.
Z'=R/2+ 2Xc so impedance neither double nor halved. So, It cannot determine without values.
Anonymous said:
9 years ago
Please Given the explanation.
ANAND said:
1 decade ago
Total impedance= R-JXc.
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