Power Dissipation in a Resistor: How Does Resistance Affect Energy Loss?

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As resistance increases in a circuit, the potential difference across the resistor also increases, initially suggesting an increase in power dissipation. However, as resistance approaches infinity, the current decreases significantly, leading to a decrease in power despite the higher voltage. The relationship between power, current, and voltage must be analyzed together to understand these dynamics. At extreme resistance values, such as 0 Ω and ∞ Ω, the behavior of current and voltage can be mathematically modeled to clarify power changes. Ultimately, as resistance increases indefinitely, power dissipation decreases due to the diminishing current.
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Homework Statement


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Homework Equations


P=V2/R
P=I2R
V=IR

The Attempt at a Solution


As the resistance of R increases, the potential difference across R increases, so the power dissipated in R increases too. But I can't seem to see how the power decreases afterwards. Is it because, as time passes, the e.m.f of the battery dies out eventually?
 
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Power is a product of current and potential difference: P = IV. What happens to the potential and current as the external resistance gets very large? Can the potential across R increase indefinitely?
 
gneill said:
Power is a product of current and potential difference: P = IV. What happens to the potential and current as the external resistance gets very large? Can the potential across R increase indefinitely?
The current becomes very small and the potential gets larger. Hmm I'm confused on which quantity to look at to determine the power.. Because if current becomes too small, then according to P=IV, power will then decrease. But as resistance increase, potential difference increases too, and again according to P=IV , power will then increase..
 
Janiceleong26 said:
The current becomes very small and the potential gets larger. Hmm I'm confused on which quantity to look at to determine the power.. Because if current becomes too small, then according to P=IV, power will then decrease. But as resistance increase, potential difference increases too, and again according to P=IV , power will then increase..
Yes, you need to consider both quantities if you want to make a choice by logical deduction. Consider V and I when R is at the extremes of its values, say 0 Ω and ∞ Ω.

Alternatively, you can analyze the circuit and write an expression for the power as a function of R, then you can examine the curve mathematically.
 
gneill said:
Yes, you need to consider both quantities if you want to make a choice by logical deduction. Consider V and I when R is at the extremes of its values, say 0 Ω and ∞ Ω.

Alternatively, you can analyze the circuit and write an expression for the power as a function of R, then you can examine the curve mathematically.
I see.. So if R -> ∞ Ω, power decreases, as current decreases right? Ok I got the picture, thanks !
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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