Hi Jut, I looked at this circuit for two days now...anyhow...if KVL is applied then don't we have to take i3 in acoount?
For loop 1 "i1" that 2Ohm resistor is also sahring some of the current of i3 isn't it???
Any suggestions anyone?
Your loop equations do not take into account the loop current I3. The equations are not quite complete. Look at the following and compare to your equations.
For I1:
80-2(I1-I3)-12I1=0
For I2:
80-8(I1+I3)-4I2=0
For I3:
-2I1+2I#+8I3+8I2+4I3=0
Simplify and get:
Equation (1) 80=14(I1) - 2(I3)
Equation (2) 80= 12(I2) + 8(I3)
Equation (3) 0=-2(I1) + 8(I2) + 14(I3)
Add equations (1) and (2) to make equation (4).
Solve Equation 1 for I3 to make equation (5).
Place the results for equation (5) into Equations (3) & (4)
Elimate (I2) by multiplying one of the equations such that the factor in fornt of I2 is equal in magnitude and opposite in sign. Add the equations together to remove I2 and solve for I1 which will equal 5 amps. V then is 5 amps X 4 Ohms = 20 volts.
Hello,
The main problem with your Loop equations is that they are not setup correctly.
You need to include the adjacent loop's affect on the Loop your writing. For Example your Loop 1. should include factor's from not only itself but the affects contributed by loops 2. & 3.
TIP:whenever a loop equation is written first include terms for the primary loop, then include the adjacent's loop's terms.
If you setup your equations properly then you should have no problem solving the corresponding Matrix.