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.
Rajesh said:
1 decade ago
For a single resistor voltage drop is maximum across that resistor, so power is maximum. But when two or more resistor come into picture then voltage drop across the resistance will be lesser than previous one so power will be less.
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.
Amarender said:
1 decade ago
In the series if we increase resistance one by one current will be lower than previous. Hence we know the P=V*I.
Or in other words if we increase the resistance power dissipation will be low according to V^2/R.
Or in other words if we increase the resistance power dissipation will be low according to V^2/R.
Ashu said:
1 decade ago
As we know current passes maximum through minimum resistive path. So as per the formula of power P=I^2*R.
Square of I decreases with increase in resistance. So minimum the resistance maximum the power.
Square of I decreases with increase in resistance. So minimum the resistance maximum the power.
Mohit said:
1 decade ago
If you know that in domestic wiring we have parallel connection to minimize power dissipation.
If we do in series then voltage will be low and power dissipation will be high.
Think about it.
If we do in series then voltage will be low and power dissipation will be high.
Think about it.
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.
VARA PRASAD said:
1 decade ago
Power consumption formula is V^2/R.
Here V is constant so whenever R value is low V^2/R value is more.
So R value is low at single 220 ohms resistor.
Here V is constant so whenever R value is low V^2/R value is more.
So R value is low at single 220 ohms resistor.
Lio said:
7 years ago
In series, current will remain the same hence;
P=i^2r.
Hence resistance directly proportional to power dissipation.
Then, How option A is correct?
P=i^2r.
Hence resistance directly proportional to power dissipation.
Then, How option A is correct?
(1)
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.
Himanish said:
1 decade ago
YES I am also having the same confusion....
P=VI
V=IR
So P = (I^2)R
BUT again P = (V^2)/R
These statements are contradictory but both are true...
P=VI
V=IR
So P = (I^2)R
BUT again P = (V^2)/R
These statements are contradictory but both are true...
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