Question

In a meter bridge circuit, the two resistances in the gap are 5 $\omega$ and 10 $\omega$. The wire resistance is 4 $\omega$. The emf of the cell connected at the ends of the wire is 5 V and its internal resistance is 1 $\omega$. What current will flow through the galvanometer of resistance 30 $\omega$ if the contact is made at the midpoint of the wire?

Answer 1

In loop $1$

$+2(I-I_1)+2(I-I_1+I_2)=55I-4I_1+2I_3=5⟶\left(1\right)$

In loop $2$

$I+5I_1+10(I_1-I_3)=5I+15I_1-10I_3=5⟶\left(2\right)$

In loop $3$

$5I_1+30I_3-2(I-I_1)=07I_1+30I_3-2I=0⟶\left(3\right)$

In loop $4$

$30I_3+2(I-I_1+I_3)-10(I_1-I_3)=042I_3+2I-12I_1=0⟶\left(4\right)$

From equatin $1$ and $2$

$26I-5I_1=30⟶\left(5\right)I_1=\frac{26I-30}{5}$

From equation $3$ and $4$

$24I-109I_1=0⟶\left(6\right)$

Using equation $5$ and $6$

$24I-109(\frac{26I}{5}-6)=0$

$24I-566.8I+654=0I=\frac{654}{542.8}=1.2$

$I_1=\frac{1.2}{5}=0.24$

Using equation $1$

$I_3=\frac{5+4I_1-5I}{2}=\frac{5+0.96-6}{2}=-\frac{0.04}{2}{I}_3=-0.02A$

Sign indicate choosen direction is opposite to the actual current direction.

Hence galvanometer read $0.02A$