# power network consumption vs network resistance [closed]

I have a simple question. Based on Ohm's law, $V=I\cdot R$, by increasing the resistance, if the voltage is fixed, the current will drop. By dropping current, the consumption or the loss will drop ($P=V\cdot I=V^2/R=R\cdot I^2$). So, the question is: does increasing total resistance of network, result in less consumption?

## closed as unclear what you're asking by Andy aka, winny, Voltage Spike, Bimpelrekkie, alex.forencichJun 7 '18 at 22:13

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• Your equations imply that $P\propto \frac{1}{R}$ when the voltage is fixed and that $P\propto R$ when the current is fixed. Beyond that, I'm not sure what your question is. – jonk Jun 2 '18 at 15:49

Based on Ohm's law, $V=I\cdot R$, by increasing the resistance, if the voltage is fixed, the current will drop.

Correct.

By dropping current, the consumption or the loss will drop ($P=V\cdot I=V^2/R=R\cdot I^2$).

Be careful with terminology.

• Consumption implies useful work done.
• Loss implies wasted energy - typically in the transmission lines or network.

Your formula shows that there is a square law at work here. At half current power is reduced to 25%.

So, the question is: does increasing total resistance of network, result in less consumption?

Now you see that your question is ambiguous. Either way, the load presented to the power supply is less so overall consumption is reduced. Whether this is good or bad depends on where the increased resistance occurs.

• My question is about the whole power network in, a country for example. "Loss" and "consumption" are both types of using electricity, aren't they? In countries around the world, people have fixed voltage at their homes, thus, if their appliances have higher resistance, will it reduce overall energy usage? – hossein Jun 2 '18 at 17:18
• "... if their appliances have higher resistance, will it reduce overall energy usage?" Yes, of course. See the last line of my answer. – Transistor Jun 2 '18 at 17:20

If you assume the voltage is constant, then yes, increasing resistance decreases dissipated power.

your reasoning is correct and this will indeed result in a lower power consumption since less current will flow.