Question

write krichhoff's law and western bridge​

Answer 1

Answer:

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Explanation:

Kirchoff's first law states that the algebraic sum of currents at a junction of an electric circuit is zero. ... Wheatstone Bridge is an electrical circuit, which is used for the accurate measurement of the resistance of a conductor.

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.

i.e. ∑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. ∑IR=∑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 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

I₁P - I₂R = 0                ........(i)

​Similarly, in the closed mesh BCDB, we have

I₁Q - I₂S = 0

I₁Q = I₂S                     ........(ii)

Dividing equation (i) by equation (ii) we get

$\frac{I1P}{I1Q} = \frac{I2R}{I2S}$    

∴ $\frac{P}{Q} = \frac{R}{S}$

This is the condition for balance of wheatstone bridge.

PLS REFER THE ATTACHED FIGURE WHILE STUDYING!!!

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