# Find he admittance in the circuit

**1. The problem statement, all variables and given/known data**

Find the admittance Y_{ab} in the circuit seen in the figure. (Units for the following are in Ohms) R1 = 5, R2 = 6, R3 = 6, R4 = 13.6. Answer in rectangular form.

**2. Relevant equations**

j = [itex]\sqrt{-1}[/itex]

Y = 1/Z

Z equivalent in series = Z_{1} + Z_{2} + … + Z_{n}

Z equivalent in parallel = [itex]\frac{1}{1/Z1 + 1/Z2 + … + 1/Zn}[/itex]

**3. The attempt at a solution**

I believe I just need to find Z_{eq.} and then take the reciprocal of that to find Y_{ab}. I’m not sure if I’m missing something, or I’m just making a calculation error.

I combined the elements in parallel first:

[itex]\frac{1}{1/(j10)+1/(6)+1/(j12+6)+1/(-j2+5)}[/itex]

Z_{parallel} = 2.51 + j*0.659

I then added the remaining impedances in series:

Z_{eq.} = -j12.8+(2.51+j*0.659)+13.6 = 16.11-j12.14

Then taking the reciprocal:

Y[ab] = [itex]\frac{1}{16.11-j12.14}[/itex] = 0.0396 + j*0.0298. I don’t know if this *exact* answer is incorrect, but I’ve put in 0.0397+j*0.3 before which is incorrect and the website usually takes answers that are very close. Any ideas? Do I have to convert everything to admittances first and then combine it or is the method I’m using correct? (EDIT) Converting everything to admittances first I got Y_{ab} = 2.05-j4.728, which I don’t know if it is correct yet. I only have one attempt remaining and don’t want to put anything in until I know.

http://ift.tt/Pszb6z

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