Prof. nchordoqui Problems set # 6 Physics 169 March 24, 2015 1. Find the current I in the circuit shown Fig. 1. 4 Ω 8 Ω 16 Ω 1 Ω 16 Ω I 3 Ω 1 Ω 20 V Figure 1: Problem 1. 2. In the circuit shown in Fig. 2, the power produced by bulb 1 and bulb 2 is 1 kw and 50 W, respectively. Which light has the higher resistance? (ssume the resitance of the light bulb remains constant with time.) 100 V bulb 1 bulb 2 i 1 i 2 Figure 2: Problem 2. 3. onsider the circuit shown in Fig. 3, with the start up switch T 1 open (for a long time). Now, close the switch and wait for a while. What is the change in the total charge of the capacitor? 4. regular tetrahedron is a pyramid with a triangular base. Six = 10.0 Ω resistors are placed along its six edges, with junctions at its four vertices, as shown in Fig. 4. 12.0-V battery is connected to any two of the vertices. Find (i) the equivalent resistance of the tetrahedron between these vertices and (ii) the current in the battery.
T 1 V 1 V 2 V 2 1 Figure 3: Problem 3. Figure 4: Problem 4. 5. Find the equivalent resistance in the limit n for the circuits shown in Figs. 5 and 6. 1 Ω 2 Ω 4 Ω 8 Ω n 1 n Figure 5: Problem 5 (i). 6. The circuit in Fig. 7 has been connected for a long time. (i) What is the voltage across the capacitor? (ii) If the battery is disconnected, how long does it take the capacitor to discharge to one tenth of its initial voltage? 7. Find the voltage across and (i.e. V ) as a function of time in the circuit shown in Fig. 8 8. The switch in Fig. 9(a) closes when c > 2/3 and opens when c < /3. The voltmeter reads a voltage as plotted in Fig. 9(b). What is the period T of the waveform in terms
37. The circuit in Figure P28.37 has been connected for a long time. (a) What is the voltage across the capacitor? (b) If the battery is disconnected, 1 Ω 2 Ωhow 3 Ωlong 4 Ω does it take n 1 n the capacitor to discharge to one tenth of its initial voltage? Figure 6: Problem 5 (ii). 10.0 V 1.00 Ω 8.00 Ω 1.00 µf µ 4.00 Ω 2.00 Ω Figure P28.37 Figure 7: Problem 6. 38. In places such as a hospital operating room and a factory for electronic circuit boards, electric sparks must be V 0 avoided. person standing on a grounded floor and touching nothing else can typically have a body capacitance of 150 pf, in parallel with a foot capacitance of Figure 8: Problem 7. 80.0 pf produced by the dielectric soles of his or her of 1, 2, and? shoes. 9. This The problemperson illustrates howacquires a digital voltmeter static affects theelectric voltage across a charge capacitor in an from circuit. digital voltmeter of internal resistance r is used to measure the voltage across a interactions with furniture, clothing, equipment, packaging capacitor after the switch in Figure P28.76 is closed. ecause the meter has finite resistance, part of materials, the current supplied by and the battery essentially passes through the everything meter. (i) pply Kirchhoff s else. rules The to thisstatic charge is conducted to ground through the equivalent resistance of the two shoe soles in parallel with each other. pair of rubber-soled street shoes can present an equiva-
68. Switch S has been closed for a long time, and the electric Problems 891 ed in paral- 1 40.0 V and charge two ing an emf. If the additional at rate does e batteries, t rate does Voltage controlled switch c (t) 2 V (a) c each other. ermine the ision of the ults in less any other in a direct r delivered to 2 3 3 T (b) Figure P28.66 Figure 9: Problem 8. t using the 28.63. The ltmeter has of actual ct to within rcuit shown 67. Three 60.0-W, 120-V lightbulbs are connected across a 120-V power source, dq as shown in Figure P28.67. Find (a) the total power eq dt delivered + q = r r + E, to the three bulbs and (b) the voltage across each. ssume that the resistance of each bulb is constant (even though in reality the resistance might increase r + markedly E 1 e t/(eq)) with q = r ( current). circuit, and use the fact that i = dq/dt to show that this leads to the differential equation where eq = r/(r + ). (ii) Show that the solution to this differential equation is and that the voltage across the capacitor as a function of time is V = r r + E(1 e t/(eq) ). (iii) If the capacitor is fully charged, and the switch is then opened, how does the voltage across the capacitor behave in this case? 10. When two slabs of n-type and p-type 1 semiconductors are put in contact, the relative affinities of the materials cause electrons to migrate out of the n-type material across the junction to the 120 V 2 p-type material. This leaves behind a volume in the n-type material 3 that is positively charged and creates a negatively charged volume in the p-type material. Let us model this as two infinite slabs of charge, both of thickness a with the junction lying on the plane z = 0. The n-type material lies Figure P28.67
differential 0 equation is increas q r # +! 28.8 (c). c r (1 e t/eq 0 0 < z < ) a % There!(x, y, z) =!(z) = $ "! 0 " a< z < 0 branch % & 0 Voltmeterz >a 28.9 (b), (d charge r a) Find the electric field everywhere. harge pacitor b) Find the potential difference between the points P 1 and P 2.. The point P 1. is located the re on a plane parallel to the slab a distance z 1 > a from the center of the slab. The charge tery vo point P 2. is located on plane parallel to the slab a distance z 2 <!a from the center of the slab. 28.10 (c), (i) on the olution: S ε circuit half of this problem, the electric field is a superposition Figure P28.76 paralle Figure 10: Problem of two 9. slabs of opposite charge density. time, t 6 fter Joseph Priest, "Meter esistance: Don't Forget It!" The Physics the rig Teacher, January 2003, p. 40. exists o utside both slabs, the field of a positive slab E! Figure 11: Problem P (due 10. to the p -type semi-conductor ) is onstant and points away and the field of a negative slab E! in the range 0 < z < a and has uniform charge density +ρ 0. The adjacent N (due p-type to the material n -type lies insemi- onductor) the is range also a constant < z < 0 and and has uniform points charge towards density ρthe 0 ; seeslab, Fig. 11. so Hence: when we add both ontributions we find that the electric field is zero outside the slabs. The fields E! P are +ρ 0 0 < z < a ρ(x, y, z) = ρ(z) = ρ 0 a < z < 0 hown on the figure below. The superposition of these fields E! T is shown on the top line 0 z > a the figure. (i) Find the electric field everywhere. (ii) Find the potential difference between the points P 1 and P 2. The point P 1 is located on a plane parallel to the slab a distance z 1 > a from the center of the slab. The point P 2 is located on plane parallel to the slab a distance z 2 < a from the center of the slab..