Emf V & Potential Difference in a Resistor

AI Thread Summary
To generate 11 W of thermal energy in a 0.31-ohm resistor connected to a 2.1 V battery, the potential difference across the resistor must be calculated using the power equation P = IV and Ohm's law V = IR. The initial calculations led to an incorrect total potential difference of 3.96 V, as the internal resistance of the battery was not considered. The correct approach involves recognizing that the voltage across the resistor is less than the battery's emf due to this internal resistance. A formula relating power, resistance, and voltage can clarify the relationship needed to solve for both the potential difference and the internal resistance. Understanding these components is crucial for accurately determining the values required in the problem.
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Homework Statement



Thermal energy is to be generated in a 0.31 resistor at the rate of 11 W by connecting the resistor to a battery whose emf is 2.1 V.

(a) What potential difference must exist across the resistor?

(b) What must be the internal resistance of the battery?


Homework Equations


P = IV
V = IR


The Attempt at a Solution


I tried manipulating the variables and substituting in the equations. P =IV, 11 = IV, V= IR, 11=I^2*R, 11=I^2*.31, so i divided 11/.31 then took the square root, which was about 6 for the current. I went back and multipled V=IR, to find V=6*.31 , which was 1.86 V. I thought that to find the total potential difference i had to add the emf V given and the V i found from current and resistance, which was 1.86V + the 2.1 V emf... 3.96. But this answer is incorrect.

As for part B, I'm completely lost.
 
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I share your confusion with this question! Only after reading part (b) do we learn that there is a second resistor involved - the internal resistance of the 2.1 Volt battery. So we don't have 2.1 Volts across the 0.31 ohm resistor. So it makes sense to ask what voltage we do have across it, as part (a) does. What do we know about this component? Resistance .31 Ohms, Power 11 Watts. We need a formula that relates V, R and P. One way to derive it is to start with P = IV and replace I with V/R, eliminating the I that we neither know or care to find.
 
Thank you so much, I was definitely over-complicating it in my thinking. Your explanation was really clear. :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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