Question

calculate the effective resistance between the point A and B in the network shown in the figure....
I want both the answer..if possible
then...u can only...give the 2nd figure tht is..the square one..plz answer ​

Answer 1

Answer:

Explanation:

FIG 1:

1 OHM AND 2 OHM ARE IN SERIES.

R SERIES = 1+2 = 3 OHMS.

THIS 3 OHM AND 1.5 OHM ARE IN PARALLEL.

R NET = 3*1.5/3+1.5

          = 4.5/4.5

          = 1 OHMS.

THEREFORE THE NET RESISTANCE IN THE FIRST CIRCUIT IS 1 OHM

FIG 2:

THIS CIRCUIT REPRESENTS A BALANCED WHEASTONE'S BRIDGE.

3/3 = 2/2 = 1.

THEREFORE NO CURRENT FLOWS THROUGH THE 1 OHM CIRCUIT

THEREFORE THE RESISTORS 3 AND 2 ARE IN SERIES IN ACB PATH AS IN SIMILAR IN PATH ADB.

R SERIES = 3+2

               = 5 OHMS.

THIS 5 OHM IN ACB AND THE OTHER 5 OHM IN ADB ARE IN PARALLEL.

R NET = 5*5/5+5

          = 25/10

          =2.5 OHMS.

THEREFORE THE NET RESISTANCE IN THE SECOND CIRCUIT IS 2.5 OHMS.

HOPE THIS HELPS.