Electrical Engineering - Series Circuits - Discussion

@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.

@Hemraj.

For P=V^2/R

In this case the resistance and power are having inverse relation, so by this we can conclude that as resistance is less power dissipation will be more.

And for P=I^2.R here the resistance and power are having direct relation, So more resistance causes of more power dissipation.

But as we know current through the CKT depends on Voltage, so we can take voltage as constant so as per P=V^2/R the resistance is less power dissipation will be more.

What about maximum power transfer theorem? Which says the max power will be transferred to the load when load impedance is equal to the source impedance. It means if load independence is greater than the source impedance then the power transferred to the load is less and vice versa. If we match both the incidences we can get max power and max heat dissipation. So the question is what is the value of source impedance of 120 volt source?

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".

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.

ANS: A

Just think when resistance keep on incresing accros the voltage source, there will not be a current at the maximum resistance, so ther will not be a power dessipation at max resistance (consider open).

Imagine when your reistance is minimum (consider short) accros voltage source, current will be high and your power desipation also will be high.

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.

V=IR
P=VI
P=V^2/R
WHEN R=220 then power will maximum
now r is denominator so if we increses r then "P" will reduces so p will maximun when r=220
now in option
anser is "A" one resistor in series so right answer
in second option two resistor in series means R increses

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.

Power= v*i =v* (v/r) = v^2 /r.

P= v^2/r it implies that the resistance and power are having inverse relation, so by this we can conclude that as resistance is less power dissipation will be more. From given options one 220 ohm resistor is less compared to other options.