Four resistors of resistance 1ohm,2ohms,3ohms,5 ohms are connected in series and afterwards in parallel. Calculate the combined resistance in each case.
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
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.
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.
(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
correction: R = 0.492 ohms.
R=R1+R2+R3+R4. =R=1+2+3+5. R=11ohms
1/R=1/R1+1/R2+1/R3+1/R4. 1/R=1/1+1/2+1/3+1/5. 61/30=2.03
PHYCIS
20
four resistors of residence are connected in series and afterwards in parallel calculate the combined resistance in each
I need a solid answer
It's ok
I need the correct answer to this question
And
0.490
Where is the 61 coming from
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