Wheatstone devised a bridge arrangement of resistances by which the resistance of a given conductor can be determined. This arrangement is called 'Wheatstone bridge'.
Arrangement
In the arrangement, four resistances are so connected as to form a parallelogram. In one diagonal of this parallelogram is connected a galvanometer and in other diagonal a cell. Now, if the resistances in the four arms of the parallelogram are so adjusted that on sending current in the bridge by the cell, there is no deflection in the galvanometer, then the bridge is said to be 'balanced'. In this condition, the ratio of resistances of any two adjacent arms of the bridge is equal to the ratio of the resistances of the remaining two adjacent arms.
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On pressing the key [tex]\sf K_2[/tex] there is no current in the arm BD. Applying kirchhoff's second law for the closed loop ABDAwe have -
Answers & Comments
Answer:
Wheatstone devised a bridge arrangement of resistances by which the resistance of a given conductor can be determined. This arrangement is called 'Wheatstone bridge'.
Arrangement
In the arrangement, four resistances are so connected as to form a parallelogram. In one diagonal of this parallelogram is connected a galvanometer and in other diagonal a cell. Now, if the resistances in the four arms of the parallelogram are so adjusted that on sending current in the bridge by the cell, there is no deflection in the galvanometer, then the bridge is said to be 'balanced'. In this condition, the ratio of resistances of any two adjacent arms of the bridge is equal to the ratio of the resistances of the remaining two adjacent arms.
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On pressing the key [tex]\sf K_2[/tex] there is no current in the arm BD. Applying kirchhoff's second law for the closed loop ABDA we have -
[tex] \sf \: I_1P - I_2R = 0 \: \: \: \: \: \: or \: \: \: \: \: \: I_1P = I_2R[/tex]
Similarly, for the closed loop BCDB, we have -
[tex] \sf \: I_1Q - I_2S = 0 \: \: \: \: \: \: or \: \: \: \: \: \: I_1Q = I_2S[/tex]
Taking them as eq (1) and (2) now divide eq (1) by eq (2) we have-
[tex] \dashrightarrow \sf \: \dfrac{I_1P}{I_1Q} = \dfrac{I_2R}{I_2S}[/tex]
[tex] \dashrightarrow \: \: \underline{ \boxed{\bf\dfrac{P}{Q} = \dfrac{R}{S}}}[/tex]
[tex] \rule{190pt}{2pt}[/tex]