Question

Resistance of four sides of Wheatstone bridge is $AB=100\Omega ,BC=10\Omega ,CD=5\Omega$ and $DA=60\Omega$. Galvanometer of $15\Omega$ resistance is connected between $BD$. Find the current flowing through the galvanometer. Potential difference between $A$ and $C$ is $10V$.

Answer 1

For closed mesh $ABDA$
$\sum E=\sum IR$
$0=100\times I_1+I_g\times 15-I_2\times 60$
$I_2\times 60=100I_1+I_{g}15$
$12I_2=20I_1+3I_g.....\left(1\right)$
For closed loop $BCDB$
$\sum E=\sum IR$
$0=(I_1-I_g)\times 10-5(I_2-I_g)-15I_g$
$0=10I_1-10I_g-5I_2-5I_g-15I_g$
$0=10I_1-5I_2-30I_g$
$5I_2=10I_1-30I_g$
$I_2=2I_1-6I_g....\left(2\right)$
Solving (1) and (2)
$I_1=\frac{25}{42}{I}_2...\left(2\right)$
In closed loop $ABCFGA$
$\sum E=\sum IR$
$10=60I_2+5(I_2+I_g)$
$10=60I_2+5I_2+5I_g$
$10=65I_2+I_g....\left(4\right)$
$2=13I_2+I_g....\left(5\right)$
Putting in equation (2)
$I_2=2I_1-6I_g$
$I_2=2\times \frac{25}{42}{I}_2-6I_g$
$0=\frac{50I_2}{42}-I_2-6I_g$
$0=\frac{8I_2}{42}-6I_g$
$0=8I_2-252I_g$
$0=4I_2-126I_g....\left(6\right)$
Solving equation (5) and (6)
$I_g=\frac{8}{1646}=0.00486A$
$=4.86\times {10}^{-3}A$.