Electrical Engineering - Series-Parallel Circuits - Discussion

Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 2)
2.
Two 1.2 k resistors are in series and this series combination is in parallel with a 3.3 k resistor. The total resistance is
138
1,389
5,700
880
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

Anand said:   3 years ago
Good explanation, thanks, everyone.

Lamba prashanth said:   4 years ago
@Deepika

You explained clearly, Thanks.
(1)

M.senthil said:   6 years ago
You are correct @Deepika.

Kumara gm said:   6 years ago
Good solution, Thanks all.
(1)

Mukesh said:   9 years ago
Good @Deepika.

R.safiyullah said:   1 decade ago
1.2+1.2 = 2.4K.

(2.4*3.3)/(2.4+3.3) = 1.38947K.

1.38947/1000 = 1.389kohm.
(2)

Md aftab alam said:   1 decade ago
Two resistors 1.2k are in series i.e 1.2k+1.2k = 2.4k ohms.

Then 3.3k ohms is parallel i.e 3.3k*2.4k/3.3k+2.4k = 1.38947k ohms.

Suresh mrcool said:   1 decade ago
Two resistors are in series.
1.2K ohm and 1.2K ohm.
Total Resistance in Series r=R1+R2=>1.2K ohm+1.2K ohm=2.4K ohm.

Then that is parrallel with 3.3K ohm.

Total resistance in Parallel=1/R=1/r+1/R3.

1/R=(1/2.4K ohm)+(1/3.3K ohm)=>R=1389.47 ohm.
(1)

Tarakant mishra said:   1 decade ago
R1=1.2,R2=1.2 they are connected in series and R3=3.3 connected in parallel with R1&R2.
R1+R2=2.4K ohm
R2parallel (R1+R2)
Formula for parallel=(R1+R2)R3/(R1+R2)+R3
=2.4*3.3/2.4+3.3
=1.38947 K ohm
=1.38947*1000=1389.47 ohms
(2)

Balakrishnanaik.b said:   1 decade ago
Good explanation deepika.


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