Question

drive the principle of whatsom bridge using kirchoffs law
?​

Answer 1

Answer:

Explanation:

Kirchhoff’s Laws

First Law

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

∑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)

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

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

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