ID: B UCSD Physics 2B Answer Section Unit Exam 3B Current, Resistance, Circuits MULTIPLE CHOICE 1. ANS: A Two ways to do this: 1. This is a compound circuit, so we must find the current around the entire loop, and therefore the equivalent resistance of the entire circuit. First, reduce the parallel leg to its equivalent. Since R 50Ω = = 25Ω . Now this is in series with R3, so N 2 = R 12 + R 3 = 25.0Ω + 50.0Ω = 75.0Ω . The total circuit current is thus V 120V I= = = 1.60A R Series 75.0Ω R 1 = R 2 , R par = R Circuit Since the entire circuit current must pass through R 3 P 3 = I 2 R 3 = (1.60A) (50.0Ω ) = 128W 2 2. Think “voltage divider”. Since R 3 = 2 R 12 , V3 = Now apply the power formula locally to R3 P3 = V 23 R3 2 = (80V ) = 128W 50Ω TOP: NET + POWER 4 2 2 V = (120V ) = 80V 3 3 ID: B 2. ANS: C Let's think of this device as being made up of many coaxial shells (layers) of length L , "width" w = 2π r and thickness dr. Since the current flows directly outward, each layer can be treated as a parallel plate resistor d whose resistance is R = ρ . The differential element for this layer is A ÊÁ d ˆ˜ dr dr dr dR = d ÁÁÁÁ ρ ˜˜˜˜ = ρ =ρ =ρ π rL A A ( r ) Lw 2 Ë ¯ Since each layer acts as one element in a series network (why?), the total resistance is the sum of these layers, that is, an integral: ∫ R = dR = ÊÁ b ˆ˜ b ρ ρ dr = ln ÁÁÁÁ ˜˜˜˜ ∫ 2π L a r 2π L Ë a ¯ Plugging in the numers, we get ÊÁ b ˆ˜ ÊÁ 2.33 ˆ˜ ρ 22Ω ⋅ m ˜˜ = 125 mΩ R= ln ÁÁÁÁ ˜˜˜˜ = lnÁÁÁ 2π L Ë a ¯ 2π (1.36m) ÁË 2.22 ˜˜¯ Note that the diameters a and b are so close that you can get an approximate answer by treating the whole device as a rolled-up plate resistor with d = b − a and A = 2π aL. Then ÊÁ 2.33 ˆ˜ ÁÁ ˜ ÁÁ 2.22 − 1 ˜˜˜ Ë d b−a b/a−1 ¯ R≈ρ =ρ =ρ = (22Ω ⋅ m) = 128mΩ A 2π aL 2π L 2π (1.36m) TOP: RESISTIVITY + INTEGRATION 3. ANS: A Begin with the definition of current density and apply the formula for drift velocity: I = JA = ÊÁË nqv d ˆ˜¯ A To find q for each particle, add up the number of charges. Neutrons have no charge, but the one proton has charge e+, Now invert the formula & solve for drift velocity vd = I 1.1 × 10 −4 A m = Ê = 7.29 × 10 7 ˆ ˆ ˆ Ê Ê 12 −3 −19 −8 2 nqA ÁÁ 41 × 10 m ˜˜ ÁÁ 1 × 1.60 × 10 C ˜˜ ÁÁ 23 × 10 m ˜˜ s ¯Ë ¯ ¯Ë Ë TOP: DRIFT VELOCITY 4. ANS: D Calculate the "time constant" and plug it into the "discharging" formula - Ê ˆÊ ˆ τ = RC = ÁÁ 250 × 10 3 Ω ˜˜ ÁÁ 20.0 × 10 −6 F ˜˜ = 5.00s Ë ¯Ë ¯ ÊÁ t ˆ˜ ÊÁ 3.00s ˆ˜ ˜˜ = 82.3V V (t) = V 0 exp ÁÁÁ − ˜˜˜˜ = (150V ) exp ÁÁÁÁ − ˜˜ ÁË τ ¯ 5.00s Ë ¯ TOP: RC TIME 2 ID: B 5. ANS: B Apply the basic formula for a resistor in cartesian (x,y,z) coordinates: R=ρ L s =ρ 2 A s Ê ˆ ⇒ ρ = Rs = (1330Ω ) ÁÁ 12.7 × 10 −2 m˜˜ = 169 Ω • m Ë ¯ TOP: RESISITIVITY 6. ANS: D Important to keep track of direction here. Since the protons and electrons are moving in the same direction, you must subtract the electron contribution to the flux since the charge is negative. Calculate the number of particles times the charge/particle and then divide by time + + − n p e + − n e e + ÊÁË n p − n e ˆ˜¯ e Q n p e + n ee I= = = = t t t t ÊÁ ˆÊ ˆ Á 6.5 × 10 17 − 3.5 × 10 17 ˜˜ ÁÁ 1.6 × 10−19 C ˜˜ Ë ¯Ë ¯ = = 0.048A = 48mA 1s TOP: CURRENT 7. ANS: D Apply Ohm’s Law: R = V 220V = = 25.1Ω I 8.75A TOP: OHM'S LAW 8. ANS: C The resistors are in parallel, so they’re at the same potential (voltage) and therefore independent of each other. That means you apply Ohm’s Law just to resistor R2 I2 = V2 V 117V = = = 6.80A R 2 R 2 17.2Ω TOP: POWER 3 ID: B 9. ANS: B Resistance is directly proportional to length. Write an equation for each wire: LA LB and R B = ρ A A RA = ρ Now divide the two equations to get the ratio formula: LA RA LA A = = RB LB LB ρ A ρ ⇒ R A = RB LA LB Since wire A is half as long as wire B, LA = 1 L 2 B ⇒ ÊÁ 1 ˆ˜ R B LA 1 LA = = R B ÁÁÁÁ ˜˜˜˜ = so that R A = R B LB LB 2 2 Ë2¯ TOP: RESISTIVITY 10. ANS: B This is basically a unit conversion problem if you think about it: energy ⇒ energy = power × time time ÁÊ ÁÊ 60 s ˜ˆ˜ J ˜ˆ ÁÊ C ˜ˆ ˜˜ E = P × t = (V × I ) × t = ÁÁÁÁ 12.0 ˜˜˜˜ ÁÁÁÁ 5.00 ˜˜˜˜ (3.00 min) ÁÁÁÁ ˜ C ¯Ë s¯ Ë Ë min ¯ Power = = 1.08 × 10 4 J = 10.8kJ TOP: ENERGY 11. ANS: A The middle lamp is "shorted out" and therefore has no current through it since the potential difference across it is zero. At the same time, the total resistance of the circuit is lowered so more current flows through the remaining bulbs. You can also see this by recognizing the circuit as a voltage divider. When the middle bulb is shorted, the battery voltage is divided between only two remaining bulbs, so they are brighter than when the voltage was divided by 3. TOP: DEMO 12. ANS: A Each differential area element dA of a circle (cross section) is a ring of thickness dr and length (circumference) 2π r. The area of this ring is the product, dA = 2π r dr ∫ I = J • dA = = 2π K ∫ R 0 J (r)2π r dr = ∫ R 0 ÊÁ 2 ˆ˜ Á Kr ˜ 2π r dr = 2π K Ë ¯ ∫ R r 3 dr 0 ÁÊ R4 1 A ˜ˆ Ê ˆ4 = (3.14) ÁÁÁÁ 122 4 ˜˜˜˜ ÁÁ 166 × 10 −2 m˜˜ = 1.45 kA 2 ¯ 4 m ¯Ë Ë TOP: CURRENT DENSITY + INTEGRATION 4 ID: B 13. ANS: B If you remembered that Q/L is charge density and speed is L/t, then ΔQ Q L Ê ˆ I= = × = λ × v = ÁÁ 2.88 × 10 −3 C / m˜˜ (26.7m / s) = 76.9 mA Δt L t Ë ¯ Even if you didn't, the only combination of the given quantities resulting in the correct units for current is the product λ × v = (linear charge density) times ( velocity) TOP: CURRENT CONCEPT 14. ANS: C This is a series combination. When all resistors are identical R SER = NR = 3 × 33Ω = 99Ω TOP: SERIES 15. ANS: C Each parallel pair has resistance R/2, so the total resistance of the circuit is R TOTAL = 3R + 3(R / 2) = 4.5R = 225Ω . Since the resistance between A and B (going around clockwise) is R AB = 2R + 2 (R / 2) = 3R, the ratio of the resistances is Hence the voltage V AB = 3R 2 = . 4.5R 3 2 (180V ) = 120V 3 Note: once you see that there are three identical sub-networks in series, the actual resistance of each element is immaterial. Just the ratio matters. You must careful: if you go the other way around, you must include the battery voltage. 16. ANS: D Definition of average current I = Q 125 C = = 7.58 A t 16.5 s TOP: CURRENT 17. ANS: D There’s only one way to do this without making yourself crazy. Make a string of coversion factors necessary to get the units right. One bulb = 30 Watts, etc etc. You have to know that June has 30 days (“30 days has September... and some months I don’t remember”) ÁÊ bulb ˜ˆ˜ ÁÊÁ 6800cents ˜ˆ˜ ÁÊÁ 1000W • hr ˜ˆ˜ ÊÁÁÁ day ˆ˜˜˜ ÁÊÁ month ˜ˆ˜ ˜˜ ÁÁ ˜˜ ÁÁ ˜˜ ÁÁ ˜˜ = 26.2 bulbs ˜˜ ÁÁÁ N = ÁÁÁÁ ˜Á ˜Á ˜ ˜ Ë 30W ¯ Ë month ¯ Ë 12cents ¯ ÁË 24hr ˜¯ Ë 30days ¯ TOP: POWER & ENERGY 5