Question

I have a physics question from my physics 102 course, and here is the question. Can someone please help? Thank You!

Find the equivalent resistance of the combination of resistors shown in the figure below. (R1 = 1.62 Mu omega, R2 = 10.6 Mu omega.)

Answer 1

8の on R 1 2 6 ημ人"deu, nesu ⑨ς.nf 8.04 2.100 μ n ala (-624.rb

Answer 2

R1 and 3.50 are in series so ,

Req 1 = 1.62 + 3.50 =5.12

now we take R2 side

0.75 and R2 are in parallel

1/Req 3 =1/10.6 + 1/0.75 after solving we get the value of Req 3

R=0.7007

R is in series with 1.50

so we get Req 3 = 1.50 +0.7007=2.2007

now we are left with 3 resistors

1) 5.12

2) 8.00

3) 2.2007

all three are in parallel so

1/Req ( final ) = 1/5.12 + 1/8 +1/2.2007

Req (final)=1.291 ----------------->>>> ANS