 #1
 20
 0
I am asked to find the total impendance ZT, voltage V2 and IL for the circuit attached
I am not sure if i am correct, but this is how i approached this problem:
So i first found the total impendance in the first parallel branch and then the second
[itex]Z=\frac{1}{\frac{1}{4<90^o}+\frac{1}{6<90^o}}=2.4<0^o[/itex]
Then i found impendance for 2nd parallel branch:
[itex]Z=\frac{1}{\frac{1}{6.8<0^o}+\frac{1}{6.8<0^o}+\frac{1}{8<90^o}}[/itex]=8.69<66.97^o
So, [itex]ZT= 8.69<66.97^o+2<0^o+2.4<0^o = 13.09<66.97^o[/itex]
Finding V2: Knowing [itex]I=4mA<0^o, Z_T=13.09<66.97^o[/itex]
[itex]E=4<0^o*13.09<66.97^o=52.36V<66.97^o[/itex]
Therefore [itex]V_2=52.36V<66.97^o[/itex]
Finding [itex]I_L[/itex]
[itex]I_L = \frac{V_L}{Z_L}=\frac{52.36<66.97^o}{8<90^o}=5.76<23.03^o mA[/itex]
I am not sure if i am correct, but this is how i approached this problem:
So i first found the total impendance in the first parallel branch and then the second
[itex]Z=\frac{1}{\frac{1}{4<90^o}+\frac{1}{6<90^o}}=2.4<0^o[/itex]
Then i found impendance for 2nd parallel branch:
[itex]Z=\frac{1}{\frac{1}{6.8<0^o}+\frac{1}{6.8<0^o}+\frac{1}{8<90^o}}[/itex]=8.69<66.97^o
So, [itex]ZT= 8.69<66.97^o+2<0^o+2.4<0^o = 13.09<66.97^o[/itex]
Finding V2: Knowing [itex]I=4mA<0^o, Z_T=13.09<66.97^o[/itex]
[itex]E=4<0^o*13.09<66.97^o=52.36V<66.97^o[/itex]
Therefore [itex]V_2=52.36V<66.97^o[/itex]
Finding [itex]I_L[/itex]
[itex]I_L = \frac{V_L}{Z_L}=\frac{52.36<66.97^o}{8<90^o}=5.76<23.03^o mA[/itex]
Attachments

41.5 KB Views: 296