Question

The current I through the network shown is​

Answer 1

Answer:

R 1 = 5ohm

R 2 =2ohm

R 3= 3 ohm

R 4= 2.5 ohm

v= 5v

i = ?

firstly by series combination ,

r= R 2 +R 3

r= (2+ 3) ohm

r= 5 ohm

by parallel combination ,

1/r1 = 1/R 1 + 1/ r

1/ r 1= (1/5 + 1/5) ohm

1/ r1 = 2/5 ohm

r1=5/2 ohm

r1= 2.5 ohm

again by series combination, we get

R = r 1+ R 4

R = (2.5+2.5)ohm

R =5 ohm

now by ohm's law, we get

v=iR

5 = i 5

i = 1A

OPTION (3) 1A IS CORRECT

HOPE IT HELPS.........