Question

Its urgent plz help me! Kindly give me right ans.only Derive principle of wheatstone bridge using kirchhoff's laws.

Answer 1

Answer:

We apply Kirchoff's current law in the shown circuit.

At junction B,

i

1

=i

g

+i

3

At junction D,

i

2

+i

g

=i

4

If current through the galvanometer is zero,

i

g

=0

thus i

1

=i

3

and i

2

=i

4

Applying Kirchoff's voltage law for loop ABDA,

i

1

P+i

g

G=i

2

R

Applying Kirchoff's voltage law for loop BCDB,

i

3

Q=i

4

S+i

g

G

When i

g

=0,

i

1

P=i

2

R

and i

3

Q=i

4

S

But i

1

=i

3

and i

2

=i

4

,

Therefore

Q

P

=

S

R