Question

With the help of Kirchhoff's Law explain the wheatstone's bridge principle.

Answer 1

Kirchhoff's law: To study complex circuits, kirchhoff's gave following too laws:
(i) If a network of conductors the algebraic sum of all currents meeting at any junction of my circuit is always zero.
i.e. $\sum I=0$
(ii) In any closed mesh (or loop) of an electrical circuit the algebraic sum of the product of the currents and resistance is equal to the total e.m.f. of the mesh.
i.e. $\sum IR=\sum E$
Formula Derivation: Referring fig.
Four resistances $P$, $Q$, $R$ and $S$ are connected to form a quadrilateral $ABCD$. A cell $E$ is connected across the diagonal $AC$ and a galvanometer across $BD$. When the current is flown through the circuit and galvanometer does not give any deflection, then the bridge is balanced and when the bridge is balanced, then
$\frac{P}{Q}=\frac{R}{S}$
This is the principle of Wheatstone bridge.
Let the current $i$ is divided into two parts $i_1$ and $i_2$ flowing through $P$, $Q$ and $R$, $S$ respectively. In the position of equilibrium, the galvanometer shows zero deflection, i.e. the potential of $B$ and $D$ will be equal.
In the closed mesh $ABDA$, by Kirchhoff's second law, we get
$i_{1}P-i_{2}R=0$
or $i_{1}p=i_{2}R$ ........(i)
Similarly, in the closed mesh $BCDB$, we have
$i_{1}Q-i_{2}S=0$
or $i_1\theta =i_{2}S$ .......(ii)
Dividing equation (i) by equation (ii) we get
$\frac{i_{1}P}{i_{1}Q}=\frac{i_{2}R}{i_{2}S}$
$\frac{P}{Q}=\frac{R}{S}$
This is the condition for balance of wheatstone bridge.