8.02X Quiz 2 Solutions

Problem 1 (25 points) Shown below is a circuit consisting of an ideal power supply with an output voltage AV and two light bulbs connected in series. Bulb 2 is seen to burn brighter than bulb 1 in this circuit (i.e. it consumes more power). Approximate the bulbs as ohmic resistances.   1 2 + I (a) In this circuit, does current flow clockwise or counterclockwise? Counterclockwise (b) Is the magnitude of the current through bulb 1 greater, the same or smaller than that of the current through bulb 2? in series: the same (c) Is the resistance of bulb 1 greater, smaller or the same as that of bulb   2? P=I²R₁ P₂ = I²R₂ P₁ < P₂ ⇒ R₁ < R₂ Smaller (d) Suppose the bulbs were connected in parallel, rather than in series. Would they burn brighter than before, dimmer than before, or the same?   Since ΔV is greater than the voltage across Bulb 1 and Bulb 2 in series, both of them will burn brighter than before. (e) Suppose the bulbs were connected in parallel, rather than in series. Would bulb 1 burn brighter, less bright, or the same as bulb 2?   [P₁ = (\frac{ΔV}{R₁})²] [P₂ = (\frac{ΔV}{R₂})²] Since R₁ < R₂ ⇒ P₁ > P₂ i.e. Bulb 1 burns brighter than Bulb 2 Problem 2 (25 points) Shown below is a graph of current vs applied voltage for a diode, like the ones used in the HVPS experiment. 2 1.5 1 0.5 0 -0.5 -1 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 [V] (a) Approximately, what is the resistance of the diode for an applied voltage of +2V? \[V = IR\] \[2V = 1A * R\] \[R = 2Ω\] (b) Approximately, what is the resistance of the diode for an applied voltage of -2V? \[V = IR\] \[-2V = 1n * R\] \[R = ∞\] (c) In one of the lecture demos, a long, resistive wire connecting a light bulb to a power supply was immersed in liquid nitrogen. Explain what happened (in a few sentences)   WIRE IS COOLED DOWN → LESS THERMAL EXCITATION OF METAL LATTICE → ELECTRONS MOVE THROUGH MORE EASILY → LESS RESISTIVITY → LESS RESISTANCE → MORE CURRENT → MORE POWER P = I²R a) [U_{o} = \frac{Q^{2}}{2C_{o}}] Change on plates remains fixed. Original energy: [U_{o} = 2J] After increase of spacing new capacitance C = ϵ₀A/(2d₀) = C₀/2 Here C₀ = ϵ₀A/d₀ ∴ U' (After increasing) = Q²/2C = Q²/2(C₀/2) = Q²/C₀ = 2 * 2J = 4J b) Remember that slab is a conductor, under these conditions the system behaves like 2 capacitors in series. Ceq | | C | | C | | | | C = 6A/(d₀/2) = 2C₀ Ceq = C₀ + C₀ = 2C₀ Ceq = 2C₀ ∴ U'(After slab inserted) = Q²/2Ceq = (Q²/2C₀) = U₀ = 2 Joules