Sol. The circuit diagram is shown in fig.

There are two nodes – G and H.

The charge flow is as indicated in the diagram

Consider the node G.

Charge entering the node = charge leaving the node.

Charge enters through AG, BG, FG and leaves through GH.

Assume the potential at the nodal point G be V G and H be V H .

Charge entering the node = 5(10 – V G ) + 6(5 – V G ) + 2(7 – V G )

= 94 – 13V G ….. (1)

Charge leaving the node = 10(V G – V H ) ….. (2)

Equating (1) and (2),

94 – 13V G = 10(V G – V H )

23V G – 10 V H = 94 ….. (3)

Consider the node H.

Charge entering the node = charge leaving the node.

Charge enters through GH, EH, CH and leaves through HD.

Charge entering the node = 10(V G – V H ) + 4(10 – V H ) + 8(12 – V H )

= 136 + 10V G – 22V H ….. (4)

Charge leaving the node = I(V H – 5) ….. (5)

Equating (4) and (5),

136 + 10V G – 22V H = I(V H –5)

23V H + 10V G = 141 ….. (6)

The number of unknowns is two and we have two equations to solve. Therefore, V G and V H can be found.

Solving the equations (3 and (6),

V G = 5.679 V

V H = 3.661 V

Charge flowing through the capacitor across
GH = C(V G – V H )

= (10 × 10 –6 ) × (5.679 – 3.661) = 2.018 × 10 –5 coulomb.

Therefore, charge flowing across the capacitor GH = 2.018 × 10 –5 Coulomb.