Electrical Engineering - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 5)
5.
Which of the following series combinations dissipates the most power when connected across a 120 V source?
one 220 
resistor
resistortwo 220 resistors
resistorsthree 220 resistors
resistorsfour 220 resistors
resistorsDiscussion:
44 comments Page 2 of 5.
Shambhavi said:
8 years ago
P=i^2 R i.e., P=V^2/R.
So if the resistance goes on decreasing, the corresponding power is less. So the minimum resistance i.e.,220 ohm delivers more power.
So if the resistance goes on decreasing, the corresponding power is less. So the minimum resistance i.e.,220 ohm delivers more power.
Angesh said:
8 years ago
Why p=i^2*r is not considered?
Vijay said:
9 years ago
Nice @Lavanya.
Susanta said:
9 years ago
When power are series it calculated as parallel so, many numbers resistor has the power are too low then the single 1/p1 + 1/p2 + 1/p3 ...< p1.
Jayaram said:
9 years ago
Nice explanation, Thank you all.
RAMESH said:
9 years ago
Nice explination @Sadik Pasha.
SHAMIK said:
10 years ago
P = V^2/R.
As R increases, P decreases.
As R increases, P decreases.
Apiaps said:
10 years ago
v = IR = I = V/R as are reduces I increases.
And P = I^2 *R.
As R reduces I increases as p increases.
And P = I^2 *R.
As R reduces I increases as p increases.
Satyaranjan acharya said:
10 years ago
Very simple thing friends P = V*I.
In this case we can increase power by two process either by increasing current or by increasing voltage though power is directly proportional to the current & voltage.
But in this question voltage is constant so we have to consider the current for solve this question. And current is higher in one resistance in comparision to series combination of 4 resistance. So the answer is "A".
In this case we can increase power by two process either by increasing current or by increasing voltage though power is directly proportional to the current & voltage.
But in this question voltage is constant so we have to consider the current for solve this question. And current is higher in one resistance in comparision to series combination of 4 resistance. So the answer is "A".
Sanoop said:
1 decade ago
P = V^2/R.
If R decreases then Power increases.
P = I^2.R.
If R decreases then current increases since I^2 is considered for power calculation, increase in current will be considered over increase in resistance.
If R decreases then Power increases.
P = I^2.R.
If R decreases then current increases since I^2 is considered for power calculation, increase in current will be considered over increase in resistance.
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