Electrical Engineering - Series Circuits - Discussion

@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.

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.

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.

Very good explanation Harpreet.

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.

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...

In series, number of resistors increase total resistance increase so resistance increase current decrease so power dessipation less.

In this case

P=i^2*r power dissipation is perportional to the ressitace. So more resistace causes of more power dissipation.

Nice explanation by lavanya.

Very good explanation Harpreet.