Physics 33: tutorial week 3,4 Capacitance, resistivity and resistance 20. How much charge flows from a 2.0V battery when it is connected to a 2.0 ñf capacitor? (24 ñc) q CV 2.0 0 6 2 2.4 0 5 C. 2. The two plates of a capacitor carry +500 ñc and 500 ñc of charge respectively, when the potential difference is 300V. Calculate the capacitance. (5 ñf) C q 500 0 6 5 0 6 F 5 ñf. V 300 22. Calculate the magnitude of the electric field between the plates of a 20 ñf capacitor if they are 2.0mm apart and each has a charge of 300 ñc. (7500Vm ) V q C 300 0 6 20 0 6 5 V. E V d 5 2 0 3 7.5 03 V m. 23. Three capacitors having capacitances of 0.6 ñf, 0.22 ñf and 0.47 ñf are connected in parallel and charged to a potential difference of 240V. (a) Determine the charge on each capacitor. (b) What is the total capacitance of the combination? (c) What is the total charge acquired? (38.4 ñc, 52.8 ñc, 2.8 ñc; 0.85 ñf; 204 ñc) (a) The charge on the 0.6 ñf capacitor q CV 0.6 240 38.4 ñc. The charge on the 0.22 ñf capacitor q CV 0.22 240 52.8 ñc. The charge on the 0.47 ñf capacitor q CV 0.47 240 2.8 ñc. (b) For a parallel combination of capacitors, C eq C + C 2 + C 3 0.6 + 0.22 + 0.47 0.85 ñc. (c) q CV 0.85 0 6 240 204 0 6 204 ñc. 24. A 6.0 ñf and a 4.0 ñf capacitor are connected in series to a 60.0V battery. (a) Calculate the equivalent capacitance. (b) What is the charge on each capacitor? (c) Determine the voltage across each capacitor. (a) For a series combination of capacitors, C eq C + C 2. Hence C C C 2 C + C 2 6.0 4.0 6.0 + 4.0 24 0 2.4 ñf,. (2.4 ñf; 44 ñc; 24V and 36V)
(b) The charge on the series combination is the same as the charge on each capacitor. Thus q CV 2.4 60 44 ñc. (c) For the 6.0 ñf capacitor V q 44 0 6 24 V. C 6.0 0 6 V q 44 0 6 36 V. C 4.0 0 6 25. Calculate the energy stored in a 600pF capacitor that is charged to 00V. (3 0 6 J) Energy 2 qv 2 CV 2 2 600 0 2 00 2 3.0 0 6 J. 26. It takes 6.0J of energy to move a 2000 ñc charge from one plate of a 5.0 ñf capacitor to the other. Calculate the charge on each plate. (0.05C) From the definition of potential difference we have V W 6.0 q 2000 0 6 3.0 03 V. The charge on each plate is therefore q CV 5.0 0 6 3.0 0 3 5 0 3 C. 27. A 6.0 ñf and a 4.0 ñf capacitor are connected in parallel and charged by a 22.0V battery. What voltage is required to charge a series combination of the two capacitors with the same total energy? (55V) The capacitor combinations both have the same energy E 2 CV 2, where C is the equivalent capacitance of the parallel or series combination. Hence 2 C parallel Vparallel 2 2 C series Vseries 2. Therefore (6.0 + 4.0) 22 2 which gives { 20 2 22 2 V 64 } 2 6.0 4.0 6.0 + 4.0 V 2, 55 V. 28. Most of the wiring in a typical house can safely handle about 5A of current. At this current level, how much charge flows through a wire in one hour? (5.4 0 4 C) I q t, q 5 3600 5.4 04 C. 29. A wire carries a current of 5A. How many electrons are flowing past any point in this wire per minute? (.9 0 2 ) I 5 A 5 C s 5 60 C min. 5 60 Hence in one minute.6 0 9.88 02 electrons pass any point in the wire. 30. Calculate the resistance of a 2.0m length of copper wire 0.5mm in diameter. Take ρ Cu.7 0 8 Ïm. (.9 Ï) R ρ l A.7 2.0 0 8 π ( 2 0.5 0 3) 2.9 Ï.
3. A wire of length 0.24m and diameter 3.0 0 5 m has a resistance of 60 Ï. Calculate the resistivity of its material. (4.7 0 7 Ïm) ρ R A l 60 π ( 2 3.0 0 5) 2 4.7 0 7 Ï m. 0.24 32. Consider a cube 5mm on a side, made of carbon. Estimate the resistance between a pair of opposite faces given ρ C 3.5 0 5 Ïm. (7 0 3 Ï) R ρ l A 3.5 0 5 5 0 3 (5 0 3 ) 2 7.0 0 3 Ï. 33. A 0.5 Ï wire is drawn out ( stretched ) to four times its original length. Assuming that the density of the wire does not change, calculate its new resistance. (8 Ï) Assume that the resistivity ρ (as well as the density) of the material is unchanged after stretching. Suppose R, l and A and R, l and A are the resistance, length and area of the sample before and after stretching respectively. After stretching, l 4 l and A 4 A, (since V l A l A 4 l A ), hence R ρ l A ρ 4 l 4 A 6 ρ l A 6 R 8 Ï. 34. A 33 Ï resistor is made from a coil of copper wire whose total mass is 2g. What is the diameter of the wire and how long is it? Take d Cu 8.9 0 3 kgm 3 and ρ Cu.7 0 8 Ïm. (0.8mm, 5.2m) Elliminate A from R ρl/a and V Al to get { } RV 2 l where V mass 2 0 3 ρ density 8.9 0 3.348 0 6 m 3. Hence { } 33.348 0 6 2 l.7 0 8 5.2 m. Using V Al and A πr 2 we can determine the diameter. The result is d.83 0 4 m. 35. A 00W light bulb has a resistance of about 2 Ï when cold and 40 Ï when on (hot). Estimate the temperature of the filament when on, assuming a mean temperature-coefficient of resistance of 6 0 3. ( 800 ) Assume that R cold R 0. Then R hot R cold ( + αt hot ) which gives T 40 2 800. 2 6 0 3 36. A coil of wire has a resistance R 0 at 0 and a temperature coefficient of resistance α. If its resistance is 20 Ï at 25 and 25 Ï at 00, calculate α and R 0. (3.64 0 3 ; 8.3 Ï) We use R T R 0 ( + αt). Thus R 25 20 R 0 ( + α 25) and R 00 25 R 0 ( + α 00). Solving the above relations for α and R 0 we find α 3.64 0 3 and R 0 8.3 Ï.
37. An iron wire has a resistance of 5.90 Ï at 20 and a gold wire has a resistance of 6.70 Ï at the same temperature. At what temperature T do the wires have the same resistance? (Take the mean temperature coefficients of resistance of iron and gold over the range from 20 to T as 5.0 0 3 and 3.4 0 3 respectively.) (66 ) We denote iron and gold by the superscripts i and g respectively. Suppose the temperature at which the iron and gold wires have the same resistance is T, then R i 0 ( + α i T ) R g 0 ( + αg T). The R 0 may be determined from R R 0 ( + αt): Solving R i 0 5.90 + 5.0 0 3 20 Ï and Rg 0 6.70 + 3.4 0 3 20 Ï. 5.90 ( + 5.0 0 3 + 5.0 0 3 T ) 20 gives T 66. 6.70 ( + 3.4 0 3 + 3.4 0 3 T ) 20 38. Three 00 Ï resistors can be connected together in four different ways, making series and/or parallel combinations. What are these four ways and what is the net resistance in each case? (33.3 Ï; 66.7 Ï; 50 Ï; 300 Ï) (a) (b) (c) (d) Each of the resistances in the diagrams (a) (c) above R 00 Ï. The equivalent resistance R eq for each of the above combinations is therefore: (a) R eq R + R + R 300 Ï, (b) R eq R + R R 0000 00 + 50 Ï, R + R 200 { } (c) R eq 00 + 66.7 Ï and 00 + 00 { (d) R eq 00 + 00 + } 33.3 Ï. 00
39. In each of the combinations below, calculate the equivalent resistance between points A and B. 60 Ω 0 Ω 8Ω (a) 50 Ω 8 Ω 5 Ω B A B 40 Ω 6 Ω (b) 3 Ω Ω 2 Ω 4Ω 6 Ω A 24 Ω B 0 Ω 8 Ω 5Ω (c) 30 Ω B (d) 3 Ω 60 40 (a) R eq 50 + 60 + 40 74 Ï. (0 + 8) 6 (b) R eq 5 + (0 + 8) + 6 9.5 Ï. (74 Ï; 9.5 Ï; 7.5 Ï; 6.8 Ï) (c) The resistance R t of the top branch is R t 8 24 + 3 + 0 Ï. Hence 8 + 24 0 30 R eq 0 + 30 7.5 Ï. (d) The 6 Ï, 5 Ï and 3 Ï resistors on the right are in series and hence have an equivalent resistance of 4 Ï. This equivalent resistance is in parallel with the 8 Ï resistor, which gives a resistance of 4 8 4+8 5.09 Ï, which in turn is in series with the 4,Ï resistor yielding 4 + 5.09 9.09 Ï. This 9.09 Ï resistance is in parallel with the 0 Ï resistor giving an equivalent resistance of 9.09 0 9.09+0 4.76 Ï in series with the 2 Ï resistor. The resistance between A and B is therefore 2 + 4.76 6.76 Ï. 40. A standard resistor marked 5 ohms is tested and found to have an actual resistance of 5.05 ohms. What length of nichrome wire of resistance 35 Ïm must be connected in parallel with the resistor to make the combined resistance of 5 ohms? (3.74 m) Let the resistance of the nichrome wire be R. R is in parallel with the 5.05 Ï resistor to give an equivalent resistance of 5.00 Ï. Hence R + 5.05, which gives R 505 Ï. 5.00 The resistance of m of nichrome wire is 35 Ï. Hence the length of nichrome wire required to have a resistance of 505 Ï is l 505 35 3.74 m.