Using nodal analysis i labeled the node closes to 2 v node v2 next v1 and the far right one vt. since the circuit is open there is no current running through the 1 ohm resistor to far right. The current value on far right says 1/2
kcl v2: (v2-2/2) + (v2-v1/2) = 1
kcl v1: (v1-v2/2) + (v1/4) +(v1-vt/2) = 0
kcl vt : (vt-v1/2) = 1/2
2((v2-2/2) + (v2-v1/2) = 1)
v2 - 2 + v2 -v1 = 2
2v2 -v1 = 2
v2 = v1/2 + 1
vt-v1 = 2
vt = 1 + v1
(4(v1-v2/2) + (v1/4) +(v1-vt/2) = 0)4
2v1 - 2v2 + v1 + 2v1 - 2vt = 0
5v1 - 2v2 - 2vt = 0
5v1 - 2(v1/2 +1) - 2(v1 +1) = 0
5v1 - v1 - 2 - 2v1 - 2 = 0
2v1 = 4
v1 = 2
vt = 1 +2
vt = 3 V
However, the answer key says that it's 4 V. Am I doing something wrong her?
voltage.jpg
(v1 - v2)/2 + (v1/4) + (v1 - vt)/2 = 0
(vt - v1)/2 = 1/2
*2: v2 - 2 + v2 - v1 = 2;
*4: 2v1 - 2v2 + v1 + 2v1 - 2vt = 0;
*2: vt - v1 = 1
v1 = 2v2 - 4;
2*(2v2 - 4) - 2v2 + 2v2 - 4 + 2*(2v2 - 4) - 2vt = 0;
vt = 1 + v1;
4v2 - 8 - 2v2 + 2v2 - 4 + 4v2 - 8 - 2*(1+v1) = 0;
4v2 - 2v2 + 2v2 + 4v2 - 8 - 4 - 8 - 2*(1+v1) = 0;
8v2 - 20 - 2 - 2v1 = 0
8v2 - 22 - 2*(2v2 - 4) = 0
8v2 - 22 - 4v2 + 8 = 0
4v2 - 14 = 0;
v2 = 3.5 v.
v1 = 2v2 - 4 = 2*3.5 - 4 = 7 - 4 = 3 volts;
vt = 1 + v1 = 4 volts;
Seems you've mistaken in arithmetics.