Question

Please solve both questions by full process

Answer 1

25. net resistance of circuit

4+6+10=20 ohm=Rnet

now battery is of 5 V

current in the circuit will be same since resistors are in series combination

I=V/R

I=5/20

I=1/4=0.25 A

now potential difference across 4 ohm=V=IR

4 ohm=0.25*4=1 V

6 ohm=0.25*6=1.5 V

10 ohm=0.25*10=2.5 V

26.Initially, for a parallel connection

1/Req = 1/R + 1/R + 1/R  (where R is the resistance of each bulb)

Req = R/3

I = 6A  

Hence V = IReq  

V = 6*R/3 = 2R

If one of the bulb is fused, then the current will still flow in the remaining 2 bulbs. The voltage will remain the same as the battery is same but the current will change.    

The new equivalent resistance would then be

1/Req = 1/R + 1/R i.e. Req = R/2

Hence I = V/Req = 2R/R/2 = 4 A

hope this helps and u understand well :)