Question

what is balance condition of Wheatstone bridge​

Answer 1

Explanation:

A Wheatstone bridge is an electrical circuit used to calculate an unknown resistance with the help of a bridge circuit.

If the value of the resistors R1, R2 and R3 are fixed, then the value of the unknown resistor RX has to be found out with the help of Kirchhoff’s voltage and current laws. For this, the bridge circuit will produce a voltage as well as a current as the circuit will not be balanced. This is the process mostly used in the measurement of strain gauges and resistance thermometers. This method is much easier to use as calculations can be done more easily with a voltage and current value than trying to make the circuit balanced.

To calculate the currents between the junctions B and D use Kirchhoff’s current law.

I3 – IX + IG = 0

I1 – I2 – IG = 0

This can be further simplified as

I3 = IX and I1 = I2…………………………………..[1]

To calculate the voltage in the loops ABD and BCD use Kirchhoff’s voltage law.

[I3 R3] – [IG RG] – [I1 R1] = 0

[IX RX] – [I2 R2] + [IG RG] = 0

On re-arranging the above equations we get the value of unknown resistance as

RX = [R2 I2 R3 I3]/ [R1 I1 IX]…………………..[2]

Substituting value of [1] in [2]

RX = [R3 R2]/ R1.

There may be cases when the values of all the resistors are known. If so, the value of the voltage between B and D [VG] can be found out from the following equation.

VG = VS * ([R3 /{R3 + RX}] – [R2/{R1 + R2}])

Answer 2

BALANCED CONDITION OF WHEATSTONE BRIDGE:

• Let's consider a WHEATSTONE BRIDGE with 4 resistances namely P, Q, R, S and a galvanometer attached in a circuit.

• Please see diagram.

Now, the balanced WHEATSTONE BRIDGE refers to a condition when NO CURRENT FLOWS THROUGH THE GALVANOMETER.

• Using Kirchhoff's Law, it can be proved that, the requirement is :

$\boxed{ \bf \: \dfrac{P}{Q} = \dfrac{R}{S}}$

This is the pre-requisites for balanced condition of the bridge.