Electrical Engineering - Series Circuits - Discussion

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.

Nice @Lavanya.

Why p=i^2*r is not considered?

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.

Clear explanation, thank you @Satyaranjan Acharya.

P is inversely proportion to R.

Nice @Lavanya.

P = V^2/R.

Choose the lowest total resistance to maximize the dissipation.

In series, current will remain the same hence;
P=i^2r.

Hence resistance directly proportional to power dissipation.

Then, How option A is correct?

@ALL.

Power, P has two formulas:
#1>> P=V^2/R and
#2>>P=I^2*R

The first relation is valid when Voltage is constant for all resistors (i.e., for Parallel combination of Rs and V.

2nd Relation is for the series circuit when Current, I is constant for all resistors.

In this question, 2nd relation is being used,
Now, by increasing Resistance, P increases, Its true But.
Value of current decreases " Square times".Hence, the Overall value of Power dissipation decrease with increase in resistance for series circuits.

I hope this will help.