Hi. Thanks for all your help. I have now solved for I as follows:
mesh 1: 120-I1(2-j5)+I2(-j5)=0
mesh 2: -14.14-j14.14-I2(-j)+I1(-j5)+I3(j4)=0
mesh 3: -j120-I3(4+j4)+I2(j4)=0
Multiplying (2) by (4+j4) and (3) by (j4)
-j113.12-I2(4-j4)+I1(20-j20)+I3(-16+j16)=0 equation (2A)...
Hello again.
gneill V3 = 20 <45 = 14.14 + j14.14 V3 is going anticlockwise and I2 clockwise hence -V3 = -(14.14+j14.14) = -14.14-j14.14 right ?
scottdave I completely forgot about converting back to polar form.
Thanks for your input guys.
Thanks for your response gneill.
I've seen the above link previously that;s why I doubted my above equations as they are different so is my whole calculation which is as follows:
Multiplying eq(2) by 4+j4 and eq(3) by j4
-113.12-I2(4-j4)+I1(20-j20)+I3(-16+j16)=0 new eq(2)...
Homework Statement
Hi guys I'm doing the mesh analysis of AC circuit and looking for some guidance.
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Homework Equations
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Good morning Everyone (if European time)
Thanking you kindly for your help. I can see the solution is relatively easy. I now plotted my graph using Excel. BvU seems that you missed v5 phase shift.
for v3 I used =SIN(3*A3*PI()/180)*28.29 , =SIN(A3*PI()/60)*28.29 works fine too.
for v5 I used...
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