The four arms of a Wheatstone bridge (Figure) have the following resistances: $AB = 100 \; \Omega$,$BC = 10 \; \Omega$,$CD = 5 \; \Omega$,and $DA = 60 \; \Omega$. $A$ galvanometer of $15 \; \Omega$ resistance is connected across $BD$. Calculate the current through the galvanometer when a potential difference of $10 \; V$ is maintained across $AC$.

(N/A) Applying Kirchhoff's loop rule to the mesh $BADB$:
$100 I_1 + 15 I_g - 60 I_2 = 0$
Dividing by $5$: $20 I_1 + 3 I_g - 12 I_2 = 0 \dots (a)$
Applying Kirchhoff's loop rule to the mesh $BCDB$:
$10(I_1 - I_g) - 15 I_g - 5(I_2 + I_g) = 0$
$10 I_1 - 10 I_g - 15 I_g - 5 I_2 - 5 I_g = 0$
$10 I_1 - 30 I_g - 5 I_2 = 0$
Dividing by $5$: $2 I_1 - 6 I_g - I_2 = 0 \dots (b)$
Applying Kirchhoff's loop rule to the mesh $ADCEA$:
$60 I_2 + 5(I_2 + I_g) = 10$
$65 I_2 + 5 I_g = 10$
Dividing by $5$: $13 I_2 + I_g = 2 \dots (c)$
Multiplying Eq. $(b)$ by $10$:
$20 I_1 - 60 I_g - 10 I_2 = 0 \dots (d)$
Subtracting Eq. $(d)$ from Eq. $(a)$:
$(20 I_1 + 3 I_g - 12 I_2) - (20 I_1 - 60 I_g - 10 I_2) = 0$
$63 I_g - 2 I_2 = 0 \implies I_2 = 31.5 I_g \dots (e)$
Substituting Eq. $(e)$ into Eq. $(c)$:
$13(31.5 I_g) + I_g = 2$
$409.5 I_g + I_g = 2$
$410.5 I_g = 2$
$I_g = \frac{2}{410.5} \approx 0.00487 \; A = 4.87 \; mA$.