Question

Derive an expression for wheatsone bridge using Kirchoff's laws.

Answer 1

Kirchhoff’s LawsFirst Law

It states that the algebraic sum of currents at a junction of an electric circuit is zero. That is,

∑I=0…(i)∑I=0…(i)

Assuming that the current flowing towards the junction as positive and that moving away from the junction as negative, then applying Kirchhoff’s first law at the junction P, we have

(+I1)+(+I2)+(−I3)or,I1+I2–I3or,I1+I2=0=0=I3…(ii)(+I1)+(+I2)+(−I3)=0or,I1+I2–I3=0or,I1+I2=I3…(ii)

Hence, the sum of currents flowing towards the junction is equal the sum of currents flowing out of the junction. This law is known as Kirchhoff’s current law.

Second Law

It states that in a closed loop of an electric circuit, the algebraic sum of emfs is equal to the algebraic sum of products of currents and the resistance in the various parts of the loop. Symbolically,

∑E=∑IR∑E=∑IR

A complex electric circuit is shown in figure. To find the current in various parts of circuit, Kirchhoff’s law can be used. Consider that the direction of emf and current flow in anticlockwise direction is taken as positive and that in clockwise direction as negative. Then applying Kirchhoff’s second law in closed loop ABCFA, we have

∑E=∑IRor,(+E1)+(−E2)or,E1–E2Similarly, in the closed loop FCDEF, we have∑E=∑IRor,(+E2)+(−E3)orE2–E3At the junction F, applying Kirchhoff’s first law, we have∑Ior,(+I1)+(+I2)+(−I3)or,I1+I2=(+I1)R1+(−I2)R2=I1R1–I2R2=(+I2)R2+(−I3)R3=I2R2–I3R3=0=0=I3∑E=∑IRor,(+E1)+(−E2)=(+I1)R1+(−I2)R2or,E1–E2=I1R1–I2R2Similarly, in the closed loop FCDEF, we have∑E=∑IRor,(+E2)+(−E3)=(+I2)R2+(−I3)R3orE2–E3=I2R2–I3R3At the junction F, applying Kirchhoff’s first law, we have∑I=0or,(+I1)+(+I2)+(−I3)=0or,I1+I2=I3

Solving these equations, we can calculate the currents I1, I2 and I3.

Wheatstone Bridge

Wheatstone Bridge is an electrical circuit, which is used for the accurate measurement of the resistance of a conductor.

In figure P, Q and R are known three resistors and X is unknown resistance. A cell E is connected across two junctions A and C and a galvanometer G between the junctions B and D. adjusting P and Q is known values, R is varied such that the current through the galvanometer is zero. So, the galvanometer shows null deflection and at this condition,

PQ=XRPQ=XR

This is called the balance condition of Wheatstone bridge. At such condition, junction B and D are at the same potential.

Verification of Balanced Condition

Suppose that the circuit was not in balanced condition initially, and the current passing through P and Q be I1 and I3, and that through X and R as I2 and I4respectively.

In the closed loop ADBA, applying Kirchhoff’s voltage law with the sign convention, we have

∑Eor,0or,I1P+IgRgSimilarly, in the closed loop BDCB, we haveor,0or,I3QAt the junction B, applying the Kirchhoff’s junction law,∑Ior,(+I1)+(−Ig)+(−I3)or,I1…(iii)and at the junction D,∑Ior,(+I2)+(+Ig)+(−I4)or,I2+Ig=∑IR=(+I2)X+(−Ig)Rg+(−I1)P=I2X…(i)=(−I3)Q+(Ig)Rg+(+I4)R=IgRg+I4R…(ii)=0=0=Ig+I3=0=0=I4…(iv)∑E=∑IRor,0=(+I2)X+(−Ig)Rg+(−I1)Por,I1P+IgRg=I2X…(i)Similarly, in the closed loop BDCB, we haveor,0=(−I3)Q+(Ig)Rg+(+I4)Ror,I3Q=IgRg+I4R…(ii)At the junction B, applying the Kirchhoff’s junction law,∑I=0or,(+I1)+(−Ig)+(−I3)=0or,I1=Ig+I3…(iii)and at the junction D,∑I=0or,(+I2)+(+Ig)+(−I4)=0or,I2+Ig=I4…(iv)

 At the balanced condition,P, Q and R are so adjusted that no current passes through the galvanometer.That is,Ig=0.At this condition, above equations change to the following forms:Ip+0×Rgor,I1PandI3Qor,I3QSimilarly,I1or,I1andI2+0or,I2=I2X=I2X=0×Rg+I4R=I4R…(vi)=0+I3=I3…(vii)=I4=I4…(viii) At the balanced condition,P, Q and R are so adjusted that no current passes through the galvanometer.That is,Ig=0.At this condition, above equations change to the following forms:Ip+0×Rg=I2Xor,I1P=I2XandI3Q=0×Rg+I4Ror,I3Q=I4R…(vi)Similarly,I1=0+I3or,I1=I3…(vii)andI2+0=I4or,I2=I4…(viii)

Dividing equation(v)by equation(vi),we getI1PI3QSubstituting the values of the current from equations(vii)and(viii),we getI3PI3Qor,PQSo, the balanced condition of Wheatstone bridge is obtained.=I2XI4R…(ix)=I4XI4R=XR

Answer 2

Entering current =leaving current