I'm trying simplify the following circuit: ( R1 in series with C1) || C2 I'm looking for Z . Thanks.
J Thread Starter jobu180 Joined Nov 1, 2008 7 Nov 1, 2008 #1 I'm trying simplify the following circuit: ( R1 in series with C1) || C2 I'm looking for Z . Thanks.
hgmjr Joined Jan 28, 2005 9,027 Nov 1, 2008 #2 HINT: Assign the series RC leg the reference Z1 and assign the C2 leg the reference Z2. Then write the expression for Z1||Z2. With the expression Z1||Z2 figured out, you can then plug in the impedances for each of the elements and solve. hgmjr
HINT: Assign the series RC leg the reference Z1 and assign the C2 leg the reference Z2. Then write the expression for Z1||Z2. With the expression Z1||Z2 figured out, you can then plug in the impedances for each of the elements and solve. hgmjr
J Thread Starter jobu180 Joined Nov 1, 2008 7 Nov 1, 2008 #3 Z1||Z2 Z1 = R1+C1 Z2 = C2 Z = Z1*Z2/(Z1+Z2) Z = (R1*C2+C1*C2)/(R1+C1+C2) Correct?
hgmjr Joined Jan 28, 2005 9,027 Nov 1, 2008 #4 jobu180 said: Z1||Z2 Z1 = R1+C1 Z2 = C2 Z = Z1*Z2/(Z1+Z2) Z = (R1*C2+C1*C2)/(R1+C1+C2) Correct? Click to expand... Sorry, No. Remember: \(\Large Z_{\small 1}=R_{\small 1}+X_{c\small 1}\) \(\Large X_{c} = \frac{1}{sC}\) hgmjr Last edited: Nov 1, 2008
jobu180 said: Z1||Z2 Z1 = R1+C1 Z2 = C2 Z = Z1*Z2/(Z1+Z2) Z = (R1*C2+C1*C2)/(R1+C1+C2) Correct? Click to expand... Sorry, No. Remember: \(\Large Z_{\small 1}=R_{\small 1}+X_{c\small 1}\) \(\Large X_{c} = \frac{1}{sC}\) hgmjr
J Thread Starter jobu180 Joined Nov 1, 2008 7 Nov 1, 2008 #5 That's right, so ... Z1 = R1 +1/sC1 Z2 = 1/sC2 Z = (R1/sC2 + 1/s^2C1C2) / (R1 +1/sC1+1/sC2) Hope that's right. Thanks.
That's right, so ... Z1 = R1 +1/sC1 Z2 = 1/sC2 Z = (R1/sC2 + 1/s^2C1C2) / (R1 +1/sC1+1/sC2) Hope that's right. Thanks.
hgmjr Joined Jan 28, 2005 9,027 Nov 1, 2008 #6 You have the the equation correct. You probably need to simplify the expression to be totally complete. hgmjr
You have the the equation correct. You probably need to simplify the expression to be totally complete. hgmjr