Question

Wheatstone bridge ABCD is arranged as follows: AB= 1 ohm, BC = 2 ohm, CD = 3 ohm, DA = 4 ohm. A resistance of 5 ohm is connected between B and D. A 4 volt battery of internal resistance 1 ohm is connected between A and C. Calculate the magnitude and direction of currents flowing through each resistance.​

Answer 1

i think ans is help you thanks

Answer 2

Answer:

Correct option is

A

0.36 A

The balanced condition for Wheatstones bridge is

Q

P

=

S

R

as is obvious from the given values.

No, current flows through galvanometer is zero.

Now, P and R are in series, so

Resistance,R

1

=P+R

=10+15=25Ω

Similarly, Q and S are in series, so

Resistance R

2

=R+S

=20+30=50Ω

Net resistance of the network as R

1

and R

2

are in parallel

i=

R

V

=

50

6×3

=0.36 A.