given a circuit powered at 12V with R1, R2, R3 respectively of 10,20,30 Ohm, determine R4 in such a way that the Wheatstone bridge is in equilibrium and then calculate voltages and currents as for reference exercise.

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.

Explanation:

higher potential

explanation:

because it is in a high speed so it must move into a of high potential