Electronics - Series-Parallel Circuits - Discussion

9. 

What is the total resistance of a circuit when R1 (7 k omega.gif) is in series with a parallel combination of R2 (20 k omega.gif), R3 (36 k omega.gif), and R4 (45 k omega.gif)?

[A]. 4 k omega.gif
[B]. 17 k omega.gif
[C]. 41 k omega.gif
[D]. 108 k omega.gif

Answer: Option B

Explanation:

No answer description available for this question.

Nandini said: (Oct 30, 2011)  
Can you explain the answer?

Nandakishor said: (Jan 12, 2012)  
Since parallel combination of resisitor is lees than least resistance . therefore parallel combination of 20 36 45 is less than 20. for this combination a series resistance of 7 is added .max answer is less than 20+7=27 there c and d options are not correct . a is also not correct because 7+anything is greater than 7.

There fore the correct answer is 17 which is option B.

Prasanna said: (Aug 19, 2012)  
RT = R1 + ( 1/R2 + 1/R3 + 1/R4)
= 7+(1/20 + 1/36 + 1/45)
=7+(10)
=17 Kohms

Revathi said: (Dec 12, 2012)  
The equivalent resistance for resistors connected in parallel is 1/Re = ((1/R2)+(1/R3)+(1/R4)).

B Nagendar said: (Feb 26, 2015)  
In parallel resistors 1/R = (1/R2+1/R3+1/R4).

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