Losing Heat

Calculus Level 1

The amount of heat dissipated through a resistor of resistance R R through a circuit with voltage V V for t t seconds is approximated by the formula H = V 2 t R H = \displaystyle\frac{V^2t}{R} .

After the circuit has been on for 5 seconds, the resistor has been heating up and its resistance is 10 Ω 10 \Omega while increasing at a rate of 1 Ω / s 1 \Omega / \text{s} . What is the rate (in J/s) that heat is being dissipated through this resistor at this moment if the circuit is operating at a constant voltage of V = 12 V = 12 ?

2.4 7.2 3.6 10.8

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Andrew Ellinor by Andrew Ellinor , 33, USA

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1 solution

Andrew Ellinor
Oct 5, 2015

Differentiate the given function with respect to time to obtain d H d t = R V 2 − V 2 t d R d t R 2 . \frac{dH}{dt} = \frac{RV^2 - V^2t\frac{dR}{dt}}{R^2}.

Substituting the values given into the result yields d H d t = ( 10 ) ( 12 ) 2 − ( 12 ) 2 ( 5 ) ( 1 ) 1 0 2 = 7.2 J/s \frac{dH}{dt} = \frac{(10)(12)^2 - (12)^2(5)(1)}{10^2} = 7.2 \text{ J/s}

2 Helpful 0 Interesting 0 Brilliant 1 Confused

The expression for H is incorrect except for the case of constant V and R. The correct expression is that the instantaneous power converted to heat is V^2/R and thus the answer should have been 14.4 J/s

Kevin Lehmann - 3 years, 8 months ago

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