You are watching: Which resistor dissipates more power?

But i thought back to messing around with resistors and also batteries, wherein I remember that if ns attached a much lower resistance come a battery, the would start to warm up much faster than a enlarge resistor. This also makes feeling in the context of the strength law, because $P=IV$ and also $I=V/R$, therefore $P=V^2/R$. So $Ppropto 1/R$ and a *smaller* resistor dissipates much more heat.

The conclusion I came to is that, in the an initial scenario, the machine in inquiry **supplies** current and also that present is *driven* across the resistor, so it"s appropriate to use $I^2 R$. However, in the 2nd scenario, $V$ is set and $I$ is determined by $R$ native there.

Is that correct?

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edited Oct 14 "14 in ~ 17:47

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asked Oct 14 "14 in ~ 17:34

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**It relies on the inner resistance of the source.**

Fist take into consideration a "voltage supply".What does "voltage supply" also mean?A voltage it is provided is claimed to calculation a fixed voltage no matter what we attach it to.Is this even possible?Suppose we affix the 2 terminals that the voltage it is provided together v a piece of wire, i.e. A yes, really low resistance pack resistor $R_L$.The existing output need to be $V/R_L$.If this is a 9 Volt battery and my resistor is 0.1 $Omega$ then I"d have actually a current of 90 Amps.A 9 Volt battery most absolutely *cannot* calculation 90 Amps.

The way to design the limitation in the battery"s maximum output current is to imagine the it is perfect voltage supply in series with an interior resistance $R_i$.Now once we connect it come a load resistor $R_L$, the full current is $I=V/(R_i + R_L)$.If $R_L ightarrow 0$ then the present goes to $I_ extmax=V/R_i$, a limited value.In various other words, the inner resistance sets a maximum calculation current.

Since the full current is $I=V/(R_i + R_L)$, the total power dissipated *in the load resistor* is

$$P_L = I^2 R_L = V^2 fracR_Lleft(R_i + R_L ight)^2.$$

In stimulate to find the value $R_L^*$ for which the strength is maximized, identify with respect to $R_L$ and collection that derivative equal to 0:

$$eginalignfracdP_LdR_L &= V^2 frac(R_i + R_L)^2 - 2 R_L(R_i + R_L)(R_i + R_L)^4\0 &= V^2 frac(R_i + R_L^*)^2 - 2 R_L^*(R_i + R_L^*)(R_i + R_L^*)^4 \2 R_L^* &= R_i + R_L^* \R_L^* &= R_i , .endalign$$

This is the an outcome to remember: *the power dissipated in the load is maximized once the pack resistance is suitable to the source"s own inner resistance*.

Now, any type of circuit you would reasonably contact a "voltage source" must have a low internal resistance compared to typical load resistances.If the didn"t climate the voltage accross the load would count on the fill resistance, i m sorry would typical your resource isn"t law a great job of gift a resolved voltage source.So, due to the fact that "voltage sources" have actually low output resistance, and because we showed that the power is maximized when the load resistance matches the resource resistance, you will certainly observe that the load gets hotter if it"s low resistance.This is why you found that v batteries, which are designed to it is in voltage sources, the lower resistance loads gained hotter.

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Current sources are the other method around.They"re designed for high inner resistance therefore you gain a hotter pack for a higher load resistance.