(a)
Nodal analysis at x
2(x - 12) + 4(x - 0) + 2(x - y) = 0
4x - y = 12 ......(1)
Nodal analysis at y
4(y - 0) + 2(y - x) = 0
3y = x ......(2)
Putting in (1)
4(3y) - y = 12
y = \(\frac {12}{11}\) volt
So,
\(V_a - V_b = \frac{12}{11} - 0\)
\(V_a - V_b = \frac{12}{11} \) volt
(b)
Emfeq = 24 - 12 = 12V
Ceq = \(\frac{2\times 4}{2 + 4} = \frac 43 \) µF
Q = CeqV
Q = \(\frac 43 \times 12\) = 16 µC
So,
\(V_a - V_b = \frac QC\)
\(= \frac{-16\mu }{2\mu }\)
\(= - 8 \,volt\)
(c)
Nodal Analysis at X
2(x - 4) + (x - 2)C + (x - 0)2 = 0
4x + Cx = 8 + 2C
x(4 + C) = 2(4 + C)
x = 2volt
So,
Va - Vb = 2 - 2 = 0 volt
(d)
Nodal analysis at Vb
(Vb - 6)4 + (Vb - 12)2 + (Vb - 24) = 0
7Vb = 24 + 24 + 24
Vb = \(\frac{72}7\)
So,
Va - Vb = 0 - \(\frac{72}7\)
Va - Vb = -10.3 volt