<***> State and explain Kirchhoff's laws How it is applicable to Wheatstone bridge ?
Kirchhoff 1st law :-
* The sum of the current flowing into a junction is equal to the sum of the current flowing away from the junction.
i1 + i3 + i5 = i2 + i4
i1+ i3 + i5 - i2 - i4 =0
Σ i = 0
*** It means the algebraic sum of the current at any junction is equal to zero.
*** It follows law of Conservation of charge.
*** Current towards junction taken as positive away from junction taken as Negitive.
Kirchhoff 2nd law (or) Loops Law :-
Any closed circuit the algebraic sum of e.m.fs and potential difference is equal to zero.
ΣE + ΣV = 0
( OR )
In any closed circuit the algebraic sum of potential difference as is equal to zero.
ΣV = 0
*** In Closed circuit ABCDA
ΣE + ΣV = 0
E - iR1 - iR2 - iR3 = 0
E = iR1 + iR2 + iR3
E = i ( R1 + R2 + R3 )
*** It follows law of conservation of energy.
Wheatstone Bridge :-
*** Wheatstone Bridge is a electrical network it works on the principle of Kirchhoff law.
*** By using this we can compare the two resistances, and to calculate unknown resistance.
*** Wheatstone Bridge consist of four resistors. P, Q, R and S are arranged on the sides of rhombus A and C are connected with battery Between B and D galvanometer ( G ) is connected as shown in the figure.
Kirchhoff 1st law apply at junction " B ".
At " B " :-
Sum of the incoming current = Sum of outgoing current
i1 = i3 + ig
* When Bridge is balanced the current flowing through the galvanometer is "zero".
ig = 0
i1 = i3 ( equation (1) )
At " D " :-
i2 + ig = i4
When Bridge is balanced " ig=0 ".
i2 = i4 ( equation (2) )
*** Kirchhoff 2nd law is Apply closed circuit " ABDA "
ΣV =0
- i1P - igG + i2 R = 0
When Bridge is balanced ig=0
-i1P + i2R = 0
i2R = i1P ( equation (3) )
** In Closed circuit " BCDB "
- i3Q + i4S + igG = 0
When Bridge is balanced ig=0
-i3Q + i4S = 0
i4S = i3Q ( equation (4) )
*** Equation (3)/ equation (4)
i2R/i4S = i1P/i3Q
P/Q = R/S
*** It is balancing condition of Wheatstone bridge.
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