The four arms of a Wheatstone bridge have the following resistances . . .

Question : The four arms of a Wheatstone bridge have the following resistances:

AB=100Ω, BC=10Ω, CD=5Ω, and DA=60Ω. 

A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Doubt by Gurdev

Solution : 

Applying Kirchhoff's Loop Rule in closed Loop 1

100I₁+15I_g-60I₂=0
5[20I₁+3I_g-12I₂]=0
20I₁+3I_g-12I₂=0/5
20I₁+3I_g-12I₂=0— (1) 

Applying Kirchhoff's Loop Rule in closed Loop 2

15I_g+5(I₂+I_g)-10(I₁-I_g)=0
15I_g+5I₂+5I_g-10I₁+10I_g=0
-10I₁+5I₂+30I_g=0 
5[-2I₁+I₂+6I_g]=0
-2I₁+I₂+6I_g=0/5
-2I₁+I₂+6I_g=0— (2)

Applying Kirchhoff's Loop Rule in closed Loop 3

60I₂+5(I₂+I_g)=10

60I₂+5I₂+5I_g=10

65I₂+5I_g=10

5[13I₂+I_g]=10

13I₂+I_g=10/5

13I₂+I_g=2 — (3)

Using Equation (1) and (2) 

20I₁+3I_g-12I₂=0

[-2I₁+I₂+6I_g=0]×10

20I₁+3I_g-12I₂=0

-20I₁+60I_g+10I₂=0

--------------------------
0+63I_g-2I₂=0

--------------------------

63I_g-2I₂=0
63I_g=2I₂
2I₂=63I_g
I₂=63I_g/2 

Putting this value of I₂ in equation (3)

13I₂+I_g=2
13[63I_g/2]+I_g=2
819I_g/2+I_g=2
409.5I_g+I_g=2
410.5I_g=2
I_g=2/410.5
I_g=20/4105
I_g=0.00487
I_g=4.87×10-3 
I_g=4.87 mA

Hence, the current through the galvanometer is 4.87 mA.