Question

answer the question fastly
WITH WORKING ​

Answer 1

Answer:

Explanation:

HERE THE NET RESISTANCE IN FIRST PARALLEL PAIR IS  

R NET₁ = 4R*4R/(4R+4R)          R {PARALLE} = R₁*R₂/(R₁+R₂)

            = 16R²/8R

            = 2 R..................(1)

THE SECOND PARALLEL PAIR

R NET₂ = 6R*12R/(6R+12R)

            = 6*12*R²/18R

            = 4R.....................(2)

NOW, 2 (R) RESISTORS R PRESENT IN BETWEEN AND ALL THESE R CONNECTED IN SERIES.

NOW NET RESISTANCE :

R NET = R NET₁+R+R NET₂+R        R {SERIES} = R₁+R₂+R₃+R₄

R NET = 2R+R+4R+R

          = 8R.

VOLTAGE APPLIED =16 V.

POWER = 4 W.

POWER = V²/R

WHERE :

P = POWER

V = VOLTAGE

R = RESISTANCE

4 = 16*16/8R

R = 16*16/4*8

R = 8 OHMS.

HOPE THIS HELPS.

Answer 2

Answer:

Correct ans is 8

I hope it's hellpfull

Plz make a brainlist