Electrical Engineering - Series-Parallel Circuits - Discussion
Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 12)
12.
The parallel combination of a 6.8 k
resistor and a 10 k resistor is in series with the parallel combination of a 2.2 k resistor and a 1 k resistor. A 100 V source is connected across the circuit. The resistor(s) with the greatest voltage drop is (are)
resistor and a 10 k resistor is in series with the parallel combination of a 2.2 k resistor and a 1 k resistor. A 100 V source is connected across the circuit. The resistor(s) with the greatest voltage drop is (are)6.8 k
2.2 k
6.8 k and 10 k
and 10 k2.2 k and 1 k
and 1 kDiscussion:
10 comments Page 1 of 1.
Naveen said:
9 years ago
V=I * R.
Are is directly proportional to we so resistance will be maximum when a voltage is maximum.
In parallel combination, voltage drop will be same so the answer is option c.
Are is directly proportional to we so resistance will be maximum when a voltage is maximum.
In parallel combination, voltage drop will be same so the answer is option c.
Bhagyashree said:
9 years ago
(6.8k * 10k)/(6.8k + 10k) = 4.047k ohms.
(2.2k * 1k)/(2.2K + 1K) = 687.5 ohms.
Now 4.047 k ohms is in series with 687.5 ohms.
FOR SERIES current is same through out the circuit so P = I^2 R.
More the R, more is the voltage drop (power dissipation) so more voltage drop in parallel combination of 6.8k and 10k.
(2.2k * 1k)/(2.2K + 1K) = 687.5 ohms.
Now 4.047 k ohms is in series with 687.5 ohms.
FOR SERIES current is same through out the circuit so P = I^2 R.
More the R, more is the voltage drop (power dissipation) so more voltage drop in parallel combination of 6.8k and 10k.
(1)
Fawad said:
10 years ago
Can someone make circuit diagram for this?
Baba said:
10 years ago
In between 6.8 K ohm and 10 K ohm which one has maximum voltage drop and how?
Jay tiwari said:
1 decade ago
@Aneesh.
In question there are two series circuit which is made by parallel combination.
The first one is 6.8 and 10 = 4.047.
Second one is 2.2 and 1 = 0.687.
We know source voltage is 100v.
So voltage division formula is v1 = (v*r1)/(r1+r2).
Now we put the value v1 = (100*4047)/4734.
V1 = 404700/4734 = 85.48v.
V2 = 100-85.48 = 14.52.
So maximum voltage drop in 'c' option.
In question there are two series circuit which is made by parallel combination.
The first one is 6.8 and 10 = 4.047.
Second one is 2.2 and 1 = 0.687.
We know source voltage is 100v.
So voltage division formula is v1 = (v*r1)/(r1+r2).
Now we put the value v1 = (100*4047)/4734.
V1 = 404700/4734 = 85.48v.
V2 = 100-85.48 = 14.52.
So maximum voltage drop in 'c' option.
Aneesh said:
1 decade ago
What will be the voltage drop against 6.8K resistor?
Gerry said:
1 decade ago
@Ramswaroop that is not allowed because the rating voltage of your is still 250 volts.
Ramswaroop Saini said:
1 decade ago
100w, 250v and 500w, 250v lamp are connected in series by 500v supply than?
Please give me answer.
Please give me answer.
Sathyanathan said:
1 decade ago
(6.8*10)/(6.8+10)=4.04
(2.2*1)/(2.2+1)=0.68
So voltage drop occur on the high resistance area connected in parallel.
(2.2*1)/(2.2+1)=0.68
So voltage drop occur on the high resistance area connected in parallel.
Chinnu said:
1 decade ago
Explanation please.
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