Current Electricity Part 2

As, A is directly connected to the positive terminal of the battery, VA=10V and VC=0
By nodal analysis at B,
‌

VB−10

100

+‌

VB−VD

15

+‌

VB−0

10

=0
53VB−20VD=30...(i)
By nodal analysis at D,
‌

VD−10

60

+‌

VD−VB

15

+‌

VD−0

5

=0
−4VB+17VD=10....(ii)
Solving Eqs. (i) and (ii) by substitution method, we get
VD=0.792V⇒VB=0.865V
The current through the galvanometer,
I=‌

VB−VD

R

Substituting the values in the above equation, we get
I=‌

0.865−0.792

15

⇒I=4.87mA