Tricky Kaplan Question! Please help!!

[4563728]

It's from the Q bank:

The equations relating to circuits are the same equations in fluids and springs. If a spring is the same as a capacitor and a dashpot (the piston that slows a door) is the same as a resistor, which system stores more energy at equal tensions, two spring with constant 3 and 4 in series or in parallel? Both systems are equidistant and the length of the stretch is the same.

A. In series
B. In parallel
C. Same
D. Depends on temperature

I picked B but thats incorrect the correct answer is C. They said the trick is that both springs have the same tensions so the force on both is the same so the stored energy so also the same, but I really don't understand their reasoning. Can some one explain this in a clearer way?

---

Members don't see this ad.

 

[Odi]

You probably picked B because total capacitance is additive for parallel capacitors (C1 + C2). But they are asking for energy..

For capacitors in parallel the energy stored is [1/2*C*V^2]. You know parallel capacitors always the same voltage on them so the total energy is proportional to the capacitance. So we can say E = C.

Lets say C1= 2uF and C2 = 4uF. Now you know that C2 will have 2x the amount of energy as C1.

Now let's change the circuit to a series.When capacitors are in series the energy stored is [1/2*(Q^2/C)]. The same current runs through a series and so the charge is equal in both capacitors. The total energy is now inversely proportional to capacitance. So we can say E = 1/C.

Lets again say C1 = 2uF and C2 = 4uF. You know that C1 will now have 2x the amount of energy as C2.

Energy is conserved.