Four resistors of resistance 1ohm,2ohms,3ohms,5 ohms are connected in series and afterwards in parallel. Calculate the combined resistance in each case.

Answer 1

This is one of the most basic setups. Clearly you need to do some reading.

series: R = 1+2+3+5 = 11
parallel:
1/R = 1/1 + 1/2 + 1/3 + 1/4
R = 12/25

Answer 2

a. R = 1+2+3+5 = 11 ohms.

b. 1/R = 1/1+1/2+1/3+1/5 = 2.033
R = 0.0492 ohms.

Answer 3

correction: R = 0.492 ohms.

Answer 4

another way to think about resistors in parallel is to add their conductances, 1/R
so total conductance = 1/1+1/2+1/3+1/5
then total resistance=1/ total conductance

Answer 5

20

Answer 6

R=R1+R2+R3+R4. =R=1+2+3+5. R=11ohms

Answer 7

1/R=1/R1+1/R2+1/R3+1/R4. 1/R=1/1+1/2+1/3+1/5. 61/30=2.03

Answer 8

PHYCIS

Answer 9

31 of May,2021
R=R1+R2+R3+R4
R=1+2+3+5= 11ohms
1/R=1/R1+1/R2+1/R3+1/R4
1/R=1/1+1/2+1/3+1/5
1/R=60/31
cross multiply
R=31/60
R=0.49ohms.

Answer 10

four resistors of residence are connected in series and afterwards in parallel calculate the combined resistance in each

Answer 11

(In series)R=R1+R2+R3+R4
R=1+2+3+5
R=11ohms
(In parallel)1/R=1/R1+1/R2+1/R3+1/R4
1/R=1/1+1/2+1/3+1/5
Finding LCM(30)
1/R=61/30
Crosss multiplying
61R=30
Dividing by 61 to get R
R=30/61
R=0.4918 or 0.492ohms

Answer 12

I need a solid answer

Answer 13

It's ok

Answer 14

I need the correct answer to this question

Answer 15

And
0.490

Answer 16

Where is the 61 coming from