Question

State Kirchhoff's law for an electrical network. Using these laws deduce the condition for balance in a Wheatstone bridge.
A wire of resistance 4R is bent in the form of a circle. What is the effective resistance between the ends of the diameter?

Answer 1

a)

Kirchhoff's law:

1) Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.

2)Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.

We apply Kirchoff's current law in the shown circuit.

At junction B,

$i_1=i_g+i_3$

At junction D,

$i_2+i_g=i_4$

If current through the galvanometer is zero,

$i_g=0$

thus $i_1=i_3$

and $i_2=i_4$

Applying Kirchoff's voltage law for loop ABDA,

$i_{1}P+i_{g}G=i_{2}R$

Applying Kirchoff's voltage law for loop BCDB,

$i_{3}Q=i_{4}S+i_{g}G$

When $i_g=0$,

$i_{1}P=i_{2}R$

and $i_{3}Q=i_{4}S$

But $i_1=i_3$ and $i_2=i_4$,

Therefore $\frac{P}{Q}=\frac{R}{S}$

b) When a wire of resistance $4R$ is bent to form a circle ,then the effective resistance across the diameter is :

$1/R_e=1/\left(2R\right)+1/\left(2R\right)=1/R$

$R_e=R\Omega$

(as the wire gets equally divided across the diameter and using wheatstone's bridge connected in parallel)