Name: Applications of Potentiometer
Uploaded: 2019-07-19
Duration: 4 min 30 s
Description: Since potentiometers are very versatile and handy to use, it is vital to understand their working principle and applications fully. In this video, learn abou...

Question

Write the principle of Wheatstone's bridge and derive the necessary condition for the balance of Wheatstone's bridge.

Accepted Answer

The Wheatstone’s bridge works on the principle that where the ratio of the resistances are equal no current flows through the galvanometer. In normal condition the galvanometer is in unbalanced condition that is current flows through the galvanometer. When no current flows through the galvanometer it is known as the balanced condition of the galvanometer.

The Wheatstone’s bridge will be in the balanced condition when the current through the galvanometer is zero.

Current will divide in magnitude to $I_{1} and I_{2}$ to go through resistors P and R.

$So I_{1} P = I_{2} R$

When no current flows through galvanometer current $I_{1}$ will go through P and Q and current $I_{2}$ will go through R and S.

So, we can write.

$I_{1} = \frac{E}{P + Q} and I_{2} = \frac{E}{R + S}$

By substituting these values in the above equation, we get that

$\frac{E}{P + Q} \times P = \frac{E}{R + S} R \frac{P}{P + Q} = \frac{R}{R + S} P (R + S) = R (P + Q) PR + PS + = RP + RQ PS = RQ This is the required condition for a balanced galvanometer . We can also write it as, \frac{P}{R} = \frac{Q}{S}$

i.e. if the ratio of the resistance connected in the circuit in this way is equal then the galvanometer will be in its balanced condition.

Write the principle of Wheatstone's bridge and derive the necessary condition for the balance of Wheatstone's bridge.

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