Electrical Engineering - Series Circuits - Discussion

Very good explanation Harpreet.

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.

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.

@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 p = v^2/r, here v constant, So pr = constant.

So for max p, r should minimum.

Resistor is a device which resists the flow of power through it, The answer will be chosen from the least value of the options simple.

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.

Power = v*2/r.

If r is minimum then power has maximum value.
And we know that i = v/r.

And p = i^2*r directionally proportional.

But at the same time Power is directly proportional to resistance only.

Higher the resistance lower the power dessipation.