If $H=\frac{V^2}{R}{t}$ ,then increasing resistance means decreasing the heat produced. But, isnt it that the heat in a circuit is produced due to the presence of resistors? Moreover metals with high resistances are used as heating elements ,like Nichrome? Why does the equation state that the heat produced is inversely proportional to Resistance
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asked Sep 6, 2016 at 11:21
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$\begingroup$ As an ignorant layman, I thought heat produced was a direct function of resistance… more resistance, more heat… Did I get even that much wrong? $\endgroup$– Robbie GoodwinCommented Dec 3 at 22:26
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6 Answers
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Your statement
If $\displaystyle{H=\frac{V^2}{R} t}$ ,then increasing resistance means decreasing the heat produced.
Implies that the voltage $V$ stays constant.
So with $V=IR$ if $V$ stays constant and the resistance $R$ increases then the current $I$ decreases.
A classic example of this happening is in a tungsten filament light bulb
When the bulb is first switched on a current flows through the filament and the power dissipated is $\displaystyle{\frac {V^2}{R}}$.
As the filament heats up the current flowing through the filament decreases because of the increased resistance of the filament and so the power dissipated decreases.
This larger current flowing through the filament is a reason why filament light bulbs often blow just as they are switched on.
Update in answer to a comment
Remembering that $V=IR$ then for a constant voltage if the resistance $R$ goes up by a factor $k$ then the current goes down by a factor $k$.
Power = $I^2 R$ so if the resistance $R$ has increased by a factor $k$ to $kR$ and the current has decreased to $\dfrac I k$ then the power is now $\left ( \dfrac {I} {k} \right )^2 kR = \dfrac {I^2R}{k}$.
This means that the electrical power dissipated has decreased by a facor $k$ when the resistance has increased by a factor of $k$.
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2$\begingroup$ That makes sense, but does that mean heat produced is due to current and not resistors? $\endgroup$ Commented Sep 6, 2016 at 12:54
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$\begingroup$ @AnamikaGhosh I have updated my answer. $\endgroup$– FarcherCommented Sep 6, 2016 at 13:16
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Agreed it is confusing, but consider two ohmic resistors, where resistor 1 is half of the value of resistor 2. And let us say that the same voltage is applied across both resistors. Now since the current through each resistor is inversely proportional to the resistance, then current in resistor 1 is twice the current in resistor 2, and the heat dissipated in resistor 1 is twice the heat dissipated in resistor 2 since heat dissipated, $H=IVt$. Same conclusion is reached by using $H=I^2Rt$
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Yes, there are two separate issues that involve resistance to keep track of. The first is, what is the current that will run through the circuit, given the voltage. That depends on the resistance such that the lower the resistance, the higher the current, and that's where the counterintuitive behavior is coming into play when you look at the heat generated. But the second question is, where is that heat generated given that you already know the current, and this is the perfectly intuitive part-- the heat comes from the highest contribution to the resistance. So the reason you get lots of heat from a smaller resister is only because the resistance in the rest of the circuit is very low-- it's no longer true if you put something in that has even less resistance than what the rest of the circuit already has.
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1$\begingroup$ Does that summarise to the fact that if a current in a circuit througha resistor is increased then the heat generated increases? $\endgroup$ Commented Sep 6, 2016 at 13:04
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1$\begingroup$ But then what is the relation between resistances and heat, because i remember learning that jeat generated by increased collision within the resistor produces heat? $\endgroup$ Commented Sep 6, 2016 at 13:05
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1$\begingroup$ *heat is generated by increased collision within the resistor $\endgroup$ Commented Sep 6, 2016 at 13:23
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$\begingroup$ Yes, heat is generated by collisions in the resistor, and yes, for a given resistor, the larger the current, the larger the heat. Also, for a given current, the heat is generated where the resistance is largest. But for a given voltage over the circuit, say from a battery, the total current can be increased by reducing the total resistance. $\endgroup$– Ken GCommented Sep 6, 2016 at 14:08
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You are just messing up with the equations. In order to state a proportionality relationship between a physical quantity and some other quantities, you need to be make sure that the quantities are all independent.
In your equation, the voltage and resistance are not independent quantities, but voltage is a function of resistance and the current flowing through it. Hence to state a relationship between resistance and heat generated, you need to crack down the voltage into independent quantities- resistance and current. Then the equation reads $\displaystyle{H=I^2Rt}$, which tells us that heat is directly proportional to resistance.
The source of your confusion was that you used voltage, which depends on current and resistance (defining relation between heat and resistance with a quantity involving resistance) to define the proportionality relation between heat and resistance.
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$\begingroup$ Is it always necessary to have independent variables before defining the relation between them? Of so, then why? $\endgroup$ Commented Sep 7, 2016 at 2:39
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$\begingroup$ Voltage and resistance are independent quantities if you do not impose restrictions on current. By "independent" is meant that each can be varied while keeping the other constant. $\endgroup$– DeepCommented Sep 7, 2016 at 5:52
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$\begingroup$ It's wrong. Resistance is a property of a given sample of certain dimensions. Current flowing through such a sample will depend on how much energy is provided for the directional motion of charges and in turn the podential drop across the sample will depend on the current flowing through it. Ohm's law treats a proportionality relation between V and I through a constant R. That's why I said mathematically voltage is a function of current and resistance. If any of the one is fixed, for example, the current, then that means voltage is a function of resistance. If resistance is fixed, V = V(I) $\endgroup$– UKHCommented Sep 7, 2016 at 13:14
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$\begingroup$ @AnamikaGhosh For example, take the case of the ideal gas equation. From that you can derive any gas law (i.e, state any relationship between the thermodynamic variables), since all the thermodynbamic variables are independent. They are mutually related to other quantities. For example pressure and volume are two degrees of freedom (two independent coordinates) of an ideal gas. But they are connected by a constraint that their product will be a constant at a definite number of particles. Like wise is Ohm's law. $\endgroup$– UKHCommented Sep 7, 2016 at 13:18
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$\begingroup$ Either you can state direct proportionality of voltage across a conductor with the current flowing through it (for a material with fixed resistance), or you can state a inverse proportionality between the current and resistance (for a constant voltage and the material's temperature varies as a function of something, like an LDR or a thermistor). But stating both at a time makes sense because current and resistance are independent and their product can be connected for a fixed resistance and current by a function called voltage. You can do the other way around too. $\endgroup$– UKHCommented Sep 7, 2016 at 13:27
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Consider analogy with fluid flow. When there flow in a pipe say, due to friction, energy of motion is dissipated away into heat. Therefore for dissipation into heat to occur two things are necessary: flow, and resistance to flow. In the absence of either of them there is no dissipation into heat. Same is true of current in a circuit.
In the equation $H=\frac{V^2}{R}t$ you can only see what the resistance is, but not what the current is. Turns out that for constant applied voltage, if you increase resistance, current reduces more than proportionately (see @Farcher's answer). In the extreme case if there is break in the circuit and therefore no current, there would be no heat generation at all.
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you all are somewhat correct if you are considering a circuit with diodes and resistors and no back flow that if you increased resistance will increase heat. . But on the flip side, lets say you connected a complete circuit with out them by connecting a wire directly from the positive side to the negative of a car battery. Then you will see the opposite reaction will happen. That a bigger shorter wire(0 gauge @ 2ft.)with less resistance will generate more heat and melt. Than if you connected a smaller longer wire(24gauge @ 50ft)with more resistance will only get hot only to a certain temperature. That it will stop generating more heat and wont melt. So actually by reducing the resistance will generate more heat quickly,and increasing resistance will lower the maximum temperature and will increase the time it takes to get to its max temp. But.. That goes along with heat dissipation based on the length or surface area,and the conductive material used, which you didn't factor.
Im not saying that decreasing the resistance reduce the total amount of heat that it will produce. Just that sometimes by reducing the resistance it can also lose its ability to dissipate heat effectively. For example.. Think of a soldering iron. That by increasing the number of turns(increase resistance) it reduces the maximum temperature it will become. But if you increase the wire size(decrease resistance)but with only enough turns to still fit inside the soldering iron. It will get hotter and be at a higher temperature than the smaller longer wire with more resistance.
Also an interesting material to check out,is vanadium oxide . Which it regulates and changes its own cooling and heating state based from outside temperature.
