KaziMahathir
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The power dissipated in a resistor is given by P = V^2/R which means power decreases if resistance increases. Yet this power is also given by P = I^2R, which means power increases if resistance increases. Explain why there is no contradiction here.
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LuigiMario
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because it starts with V/IR triangle

thats

V
---
I*R

and the 'second derivative' equations are just variations on V=I*R, R=V/I, I=V/R

so Power = Vsquared divided by R, is a reasonable equation, if Volts goes up a bit (and resistance is static) then power will increase due the square of the voltage.

whilst Power = I(current)squared times resistance, if current goes up a bit & (and resistance is static) then power will increase due the square of the current,
but if you consider, as your question asked, what if resistance increases?, in most simple circuits, when the resistance increases the current decreases!
Everything in the V/IR triangle is stable at a particular time, if R increases, usually someone has changed the circuit (*)

Ohms law, where it applies, is your friend


(*) there are VDR voltage dependent resistors & temperature dependent resistors = thermometers, but you often aren't asked about these in Power questions. In fact all resistors are a little bit temperature dependent , they have a TCR, temperature coefficient of resistance, again not usually dealt with at this level. This was the basis of the BBC Engineering recruitment interview, show a simple circuit, get candidate to describe first level & approximations, then move into real world second & further derivatives, down to atomic properties of the electronic materials. Presumably they only need "talent" nowadays
Last edited by LuigiMario; 1 month ago
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Joinedup
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I, V and R are related by the formula V=IR

You can't vary them all independently to any arbitraty value you like.

If you've got a constant voltage source the current must go up as the resistance goes down.
Because power is volage times current and the voltage is constant but the current increased the power increases.
---
If you've a constant current source the voltage goes down as the resistance goes down. The power goes down because power is voltage times current.

----
Most of the sources you'll be looking at will be constant voltage (or approximate constant voltage for the sake of the pedants gallery)

Things such as batteries, mains voltage, regulated low voltage power supplies plugged into mains etc are all (approximately) constant voltage
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KaziMahathir
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(Original post by LuigiMario)
because it starts with V/IR triangle

thats

V
---
I*R

and the 'second derivative' equations are just variations on V=I*R, R=V/I, I=V/R

so Power = Vsquared divided by R, is a reasonable equation, if Volts goes up a bit (and resistance is static) then power will increase due the square of the voltage.

whilst Power = I(current)squared times resistance, if current goes up a bit & (and resistance is static) then power will increase due the square of the current,
but if you consider, as your question asked, what if resistance increases?, in most simple circuits, when the resistance increases the current decreases!
Everything in the V/IR triangle is stable at a particular time, if R increases, usually someone has changed the circuit (*)

Ohms law, where it applies, is your friend


(*) there are VDR voltage dependent resistors & temperature dependent resistors = thermometers, but you often aren't asked about these in Power questions. In fact all resistors are a little bit temperature dependent , they have a TCR, temperature coefficient of resistance, again not usually dealt with at this level. This was the basis of the BBC Engineering recruitment interview, show a simple circuit, get candidate to describe first level & approximations, then move into real world second & further derivatives, down to atomic properties of the electronic materials. Presumably they only need "talent" nowadays
That clears things up! Thank you.
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