Question

Using Kirchhoff’s rules, obtain the balance condition in terms of the resistances of four arms of wheatstone bridge.

Answer 1

Wheatstone Bridge Principle : 

It states that if four resistances P,Q,R and S are connected to form a bridge and bridge is balanced 

(i.e … if galvanometer shows no deflection), 

Then,

Proof :

Applying kirchhoff’s loop rule in ABDA,

I₁P + I_gG – (I - I₁)R = 0 …………(1) 

G : resistance of galvanometer 

Now, 

Applying kirchhoff’s loop rule in BCDB, 

(I₁ - I_g)Q – (I - I₁ + I_g)S -I_gG = 0 ………(2)

‘R’ is so adjusted such that Ig = 0.

Putting Ig = 0 in (1) & (2),

I₁P – (I – I₁)R = 0 

Or, I₁P = (I – I₁)R …….(3) 

And,

I₁Q – (I - I₁)S = 0 

⇒I₁Q = (I – I₁)S……(4) 

On dividing (3) and (4) , we have :