Question

Explain wheatstone bridge on the basis of kirchoffs law
pls give relevant answer​

Answer 1

Answer:

QP=SR. This is the principle of Wheatstone bridge. Let the current i is divided into two parts i1 and i2 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.

congrats for ur genius rank

Answer 2

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.

(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.

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

P/Q = R/S

This is the principle of Wheatstone bridge.

Let the current i is divided into two parts i¹ and i² 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...(Rest given in the attachment)