Question

Derive the condition for balance of a Wheatstone’s network using Kirchhoff’s laws. Name a device which works on the principle of balanced Wheatstone’s network.
(and tell me from which chapter the question is)​

Answer 1

Answer:

Wheatstone bridge (balanced) - Let i be the current from battery E .At point A , current i  

Dividing eq1 from eq2 , we get

        P/Q=R/S    

This is the condition for balance in a Wheatstone's bridge

Explanation:

i hope this helps u

Answer 2

Answer:

Based on the circuit given in the figure , we have three known resistance's R1, R2, R3 and an unknown variable resistor Rx, a source of voltage, and a sensitive ammeter. Kirchhoff's first rule is applied to find the currents in junctions B and D:

I3 - Ix + Ig = 0

I1 - I2 - Ig = 0

Now Kirchoff's second rule is used to find the voltage in the loops ABD and BCD:

I3 .R3 − Ig .Rg − I1.R1 = 0

Ix.Rx − I2.R2 + Ig.Rg = 0

The bridge is balanced and Ig = 0, so the second set of equations can be rewritten as:

I3.R3 =I1.R1

Ix.Rx = I2.R2

I3 =Ix

I1=I2

13.R3/1x.Rx = I1.R1/I2.R2

R1/R2 = R3/Rx.

The Wheatstone bridge is a device which works on the principle of balanced Wheatstone’s network.