Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 15)
15.
The parallel combination of a 470 
resistor and a 680 resistor is in series with the parallel combination of four 2 k resistors. The total resistance is
resistor and a 680 resistor is in series with the parallel combination of four 2 k resistors. The total resistance is1,650
1,078
77.8
778
Discussion:
14 comments Page 1 of 2.
Ravi said:
1 decade ago
First parallel combination of a 470 ohm resistor and a 680 ohm resistor = 277.91 ohm (w.r.t (r1*r2)/(r1+r2))
Second parallel combination of four 2 k ohm resistors = 0.5 K ohm = 500 ohm
Both first and second parallel combinations are in series to each other hence total resistance will be:
277.91 ohm + 500 ohm = 777.91 ohm i.e. 778 ohm
Second parallel combination of four 2 k ohm resistors = 0.5 K ohm = 500 ohm
Both first and second parallel combinations are in series to each other hence total resistance will be:
277.91 ohm + 500 ohm = 777.91 ohm i.e. 778 ohm
SANDEEP K PATIL said:
8 years ago
Second parallel combination is very simple, if the number of resistors are connected of same value then the simple formula is there resistance value/ number of resistors.
From above problem, it has four, 2K ohm resistors connected in parallel then= 2000/4= 500 ohm.
From above problem, it has four, 2K ohm resistors connected in parallel then= 2000/4= 500 ohm.
Prabhakaran said:
9 years ago
1st parallel combination = (470 * 680/470 + 680) = 277.91,
2nd parallel combination,
Step1: (R1 * R2/R1 + R2) = 1,
Step2: (1 * R3/1 + R3) = 0.666,
Step3: (0.666 * 2/0.666 + 2) = 0.500 k ohm = 500 ohm,
Req = 277.91 + 500 = 777.91.
2nd parallel combination,
Step1: (R1 * R2/R1 + R2) = 1,
Step2: (1 * R3/1 + R3) = 0.666,
Step3: (0.666 * 2/0.666 + 2) = 0.500 k ohm = 500 ohm,
Req = 277.91 + 500 = 777.91.
(1)
Saurabh K. Chhaya said:
1 decade ago
R1 (1st parallel comb. of 470 & 680)=333.33
R2 (2nd parallel comb. of 4 nos. of 2000) =.002
Total R (series comb. of R1 & R2) = 333.332 OHMS
Please check this with answer.
R2 (2nd parallel comb. of 4 nos. of 2000) =.002
Total R (series comb. of R1 & R2) = 333.332 OHMS
Please check this with answer.
Mahi said:
7 years ago
1st parallel combination:
470*680/470+680
470*680/1150
277.91.
2nd series & parallel combination.
2kohm=2000ohm,
2000/4=500,
Total R= 277.91+500=777.91Ω,
i.e. 778Ω.
470*680/470+680
470*680/1150
277.91.
2nd series & parallel combination.
2kohm=2000ohm,
2000/4=500,
Total R= 277.91+500=777.91Ω,
i.e. 778Ω.
(3)
Mujib said:
5 years ago
@Riya,
Solving first parallel combination becomes 277.91.
Because R1 multiply by R2 divide by R1+R2.
470 * 680/470 + 680 = 319600/1150 = 277.913.
Solving first parallel combination becomes 277.91.
Because R1 multiply by R2 divide by R1+R2.
470 * 680/470 + 680 = 319600/1150 = 277.913.
Daling said:
1 decade ago
4 nos of 2k ohms are parallel that's why
2*2/4=1 and 2*2/4=1
So
1*1/2=0.5k ohms=500 ohms
Ok I hope you can understand my dear jiban.
2*2/4=1 and 2*2/4=1
So
1*1/2=0.5k ohms=500 ohms
Ok I hope you can understand my dear jiban.
Riya said:
6 years ago
First parallel combination 470 and 680 is 2779.1.
Not 277.91.
Correct me, if I am wrong.
Not 277.91.
Correct me, if I am wrong.
Ravi said:
1 decade ago
= Parallel + series + parallel.
= (470*680/470+680)+(2000/1+1+1+1).
= 778 answer.
= (470*680/470+680)+(2000/1+1+1+1).
= 778 answer.
Chaithra said:
8 years ago
I got the second step easily from your explanation. Thank you @Sandeep.
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