Electrical Engineering - Series Circuits - Discussion

1. Since voltage is constant (120 v) and P = V^2/R which implies that power is inversely proportional to resistance. henceforth power will dissipate more when resistance is least i.e option A.

OR
2. P = I^2R. here power depends on the square of the current and current is inversely proportional to the current(ohm's law). I = more when resistance is less hence more current = more power PERIOD.

See.

When any resistor connected in series combination, then. Power is directly proportional to Current (I). So, A/q to find current _we have voltage and resistance.

Therefore, P=[I]2 (square)R. So, when we find the value of current, we will see that (a) option have a maximum value.

Keeping voltage constant power is inversely proportional to resistance, so resistance increases power decrease.

For an Electrical circuit.

P=I*I*Req,
Req= equivalent resistance of the circuit.
We can also write it as P=V*V/Req.

Now
case1- one 220ohm resistance Req=220ohm,
case2-Two 220ohm resistance.
Req=2*220ohm ,P2=P1/2,
Similarly P3=P1/3.
P4=P1/4.
Hence max power is dissipated in the case 1.

@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.

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 = V^2/R.

Choose the lowest total resistance to maximize the dissipation.

Nice @Lavanya.

P is inversely proportion to R.

Clear explanation, thank you @Satyaranjan Acharya.