q6_sol_3-5 - ΔU(bat =-V ΔQ =(1/6)CV 2 ΔU(bat(U’-U...

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Quiz 6- 3,4,5 C C’ V = V = V C S’ C’=2C C P’ = C+C’ = 3C 1/C S’ = 1/3C+1/3C = 2/3C C S’ = (3/2)C U’ = C S’ V 2 /2 = (3/4)CV 2 U’-U = (3/4 – 2/3)CV 2 = (1/12)CV 2 ΔQ ≡ (Q S’ – Q P ) = (C s’ - C P ) V = (3/2-4/3)CV = (1/6)CV >0 Potential difference across battery is constant, V, so battery lost charge and thus energy
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Unformatted text preview: ΔU(bat) = -V ΔQ = -(1/6)CV 2 ΔU(bat)/(U’-U) = -(1/6)/(1/12) = -2 < -1. False: E(tot) always conserved (wires were heated by current that flowed while charges were rearranging). Switch closed C’ C C P’ C P’...
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