
#1
Oct1611, 05:31 PM

P: 13

1. The problem statement, all variables and given/known data
In the figure, calculate v_0 and i_0 when v_s = 1 2. Relevant equations Using Mesh Currents: KVL: The sum of all voltages around a closed loop is equal to zero. Ohm's Law: V = IR Cramer's Rule: i2 = delta2/delta 3. The attempt at a solution I first designated the lower left loop as i1, the lower right loop as i2 and the top loop as i3 (All going counter clockwise). I noted that v_s = 1 and i0 = i2 Using KVL, I found the equations for each mesh. Mesh 1: v_s + 1(i1i3) + 1(i1i2) = 0 2*i1  i2  i3 = v_s 2*i1  i2  i3 = 1 Mesh 2: 1*i2 + 1(i2i1) + 1(i2i3) = 0 i1 + 3i2  i3 = 0 Mesh 3: 1*i3 + 1(i3i2) + 1(i3i1) = 0 i1  i2 + 3*i3 = 0 So I have my three equations and three unknowns. I also know that v0 is the voltage drop across the 1 ohm resistor, which is v0 = ir = i2*1 = i2 and the current i0 is the same as i2 since it is in the same loop i0 = i2. So I create a matrix with the three equations and attempt to solve the matrix using determinants and Cramer's rule. [ 2 1 1 ] [i1] = [1] [ 1 3 1 ] [i2] = [0] [ 1 1 3 ] [i3] = [0] delta = det(A) = 2(91) + 1(31)  1(1+3) = 10 delta2 = det(A (with 2nd column as 1, 0, 0)) = 2*0  1(3+1) + 1*0 = 2 So i2 = delta2/delta = 2/10 = .2 Amps v0 = i2*1 = .2 Volts i0 = i2 = .2 Amps I checked through my equations and these seems like the correct answers, but the solution says that it is .5 amps and .5 volts. Any insight on why the answer would be different would be helpful. 



#2
Oct1611, 07:22 PM

Mentor
P: 11,416

det(A) is not 10, and the determinant of your adjoint matrix for i2 is not 2. Better check your matrix math!




#3
Oct1611, 08:56 PM

P: 13

Wow I fail at adding and subtracting. Thanks for your input, got the correct determinants now.



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