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in Class 12 by kratos

State Kirchoff'* law for an electrical network. Using these laws deduce the condition for balance in a Wheatstone bridge. deduce the condition for balance in a Wheatstone bridge.

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+1 vote
by kratos
 
Best answer

*(1) Kirchhoff' first law (Junction rule or KCL) :**The algebraic sum of the currents at any junction is zero.

∴ ∑I = 0

(or)

The sum of the currents flowing towards a junction is equal to the sum of currents away from the junction.

*(2) Kirchhoff' second law (Loop rule or KVL) :** The algebraic sum of potential around any closed loop is zero.

∴ ∑(IR) + ∑E = 0

Wheatstone bridge : Wheatstone'* bridge circuit consists of four resistances R1, R2, R3 and R4 are connected to form a closed path. A cell of emf ε is connected between the point A and C and a galvanometer is connected between the points B and D as shown in fig. The current through the various branches are indicated in the figure. The current through the galvanometer is Ig and the resistance of the galvanometer is G.

Applying Kirchhoff'* first law

at the junction D, I1 – I3 – Ig = 0 ..... (1)

at the junction B, I2 + Ig – I4 = 0 ..... (2)

⇒ Applying Kirchhoff'* second law to the closed path ADBA

–I1R1 –IgG + I2R2 = 0 or

⇒ I1R1 + IgG = I2R2 ..... (3)

⇒ to the closed path DCBD

–I3R3 + I4R4 + IgG = 0

⇒ I3R3 – IgG = I4R4......(4)

⇒ When the galvanometer shows zero deflection the points D and B are at the same potential so Ig = 0.

Substituting this value in (1), (2), (3) and (4).

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