Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Galvanometer wikipedia , lookup
Giant magnetoresistance wikipedia , lookup
Rectiverter wikipedia , lookup
Nanogenerator wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Rechargeable battery wikipedia , lookup
Negative resistance wikipedia , lookup
RLC circuit wikipedia , lookup
Two-port network wikipedia , lookup
Lumped element model wikipedia , lookup
Electricity Review Sheet 1 Consider the following statements given below and determine if the charge of Object A is: A. positive B. negative C. neutral I. Object A is charged by friction using animal fur. Animal fur has a weaker electron affinity than object A. II. Object A is charged by contact using a negatively charged object. III. Object A is charged by induction using a positively charged object. IV. Object A is attracted to a negatively charged object. 2 A glass rod is given a positive charge by rubbing it with silk. The rod has become positive by (A) losing electrons (C) losing protons (B) gaining electrons (D) gaining protons 3 A positively charged rod is held near the knob of a neutral electroscope. Which diagram best represents the distribution of charge on the electroscope? 4 9 The diagram below represents the electric field surrounding two charged spheres, A and B. What is the sign of the charge of each sphere? (A) Sphere A is positive and sphere B is negative. (B) Sphere A is negative and sphere B is positive. (C) Both spheres are positive. (D) Both spheres are negative. 5 What is the magnitude of the electrostatic force between two electrons separated by a distance of 1.00 × 10–8 meter? (A) 2.56 × 10–22 N (C) 2.30 × 10–12 N –20 (B) 2.30 × 10 N (D) 1.44 × 10–1 N 6 A negatively charged comb has –8.0 C of charge on it. How many excess electrons are on it? (A) 5 × 1013 (C) 2 × 10-14 19 (B) 5 × 10 (D) 5 × 10-25 7 The electrical resistance of a metallic conductor is inversely proportional to its (A) temperature (C) length (B) cross-sectional area (D) resistivity 8 What is the current in a 100-ohm resistor connected to a 0.40-volt source of potential difference? (A) 250 mA (C) 2.5 mA (B) 40 mA (D) 4.0 mA The potential difference between two points in an electric field is 10 volts. The energy to move a positive charge of 2 × 10-4 C from a point at the lower potential to a point at the higher potential is: (A) 5 × 104 J (C) 2 × 10-3 J (B) 2 × 10-2 J (D) 2 × 10-5 J 10 At a distance 10 cm from a point charge, the electric field is 3600 N/C and points toward the charge. The amount of this point charge is (A) 40 microcoulombs (B) 4 microcoulombs (C) 4 nanocoulombs (D) 4 megacoulombs 11 The diagram to the right shows a point, P, located between two oppositely charged parallel plates. The electric field between the plates (A) points up, and is varying (non-uniform) (B) points down, and is varying (non-uniform) (C) points up, and is constant (uniform) (D) points down, and is constant (uniform) 12 The diagram to the right represents an electric circuit consisting of a 12-volt battery, a 3.0-ohm resistor, R1, and a variable resistor, R2. At what value must the variable resistor be set to produce a current of 2.0 ampere through R1? (A) 6.0 Ω (C) 3.0 Ω (B) 9.0 Ω (D) 12 Ω 13 An electric iron operating at 120 volts draws 10 amperes of current. How much heat energy is delivered by the iron in 30 seconds? (A) 3.0 × 102 J (C) 3.6 × 103 J (B) 1.2 × 103 J (D) 3.6 × 104 J 14 A 20-ohm resistor and a 30ohm resistor are connected in parallel to a 12-volt battery as shown. As shown to the right, an ammeter is connected. What is the equivalent resistance of the circuit? (A) 10 Ω (C) 25 Ω (B) 12 Ω (D) 50 Ω 15 In the previous question, what is the current reading of the ammeter? (A) 1.0 A (C) 0.40 A (B) 0.60 A (D) 0.20 A 16 In the previous question, what is the power of the 30-ohm resistor? (A) 4.8 W (C) 30 W (B) 12 W (D) 75 W 17 Two resistors are connected to a source of voltage as shown in the diagram below. At which position should an ammeter be placed to measure the current passing only through resistor R1 and a voltmeter be placed to measure the potential difference across R2, respectively? (A) 2 and 1 (C) 3 and 1 (B) 3 and 2 (D) 4 and 1 21 The schematic diagram to the right shows and ammeter A, and three resistors. Solve the following with equations, then check with tables: a Calculate resistance R1. b Calculate current in R2. c Calculate current in R3. d Calculate power in R1. 22 The wiring diagram shown below shows four light bulbs and a battery arranged in a combination circuit. The battery is 6 volts, and each light bulb is 12 . 18 Which combination of resistors has the smallest equivalent resistance? Problem Solving: Do the following problems on separate paper. Show the general equation, substitutions, calculations, final results with units. 19 The drawing below shows the atomic nucleus of two atoms each with 8 protons, and a nearby electron. 2.4 x 10-14 electron a b c 23 The schematic diagram shown below shows five resistors and a battery are arranged in a combination circuit. Solve with tables. 5.9 x 10-15 protons Find the force between the two groups of protons. Find the net force on the electron. Find the electric field at the electron’s location. 20 Two resistors are wired in series with a 12-volt battery. The first resistor has 10 volts across it, and the second resistor has 0.5 amps through it. Solve the following with equations, then check with tables: a b c d e a Draw the schematic diagram for the circuit. b Calculate the circuit’s equivalent resistance. Use tables to find: c Calculate the potential difference of each resistor. d Calculate the current through each resistor. e Calculate the power consumed by each resistor. Draw and label the circuit schematic diagram. Calculate the potential difference across R2. Calculate the resistance R1. Calculate the combined resistance of R1 and R2. Calculate the amount of power R1 consumes. a b Calculate the equivalent resistance, total current, and total power for this circuit. HONORS: Calculate the current, voltage, and power of all five resistors. 24 A +3.0 nanocoulumb charge is located 1.0 m to the left of a –1.0 nanocoulumb charge. a b c Draw the field lines around these two charges. Calculate the electric field at the midpoint between the two charges. HONORS: Where should a third charge of +2 nC be placed so that it is in static equilibrium? Answers 1 I-B, II-B, III-B, IV-A or C 2 A 3 D 4 B 5 C 6 A 7 B 8 D 9 C 10 C 11 D 12 C 13 D 14 B 15 B 16 A 17 C 18 D 19a 424 N b 5.26 N, right c 3.29 × 1019 N/C, left 20b 2 v c 20 Ω d 24 Ω e 5 W 21a 200 Ω b 0.5 A c 0.67 A d 72 W 22b 20 Ω c VA–VD : 3.6, 1.2, 1.2, 2.4 V d IA–ID: 0.3, 0.1, 0.1, 0.2 A e PA– PD : 1.08, 0.12, 0.12, 0.48 W 23a 12 Ω, 3 A, 108 W b I1–I5: 1.8, 1.8, 1.2, 3, 3 AV1–V5 : 10.8, 7.2, 18, 12, 6 V; P1– P5 : 19.44, 12.96, 21.6, 36, 18 W Electricity Review Sheet Solutions 1. answers: I – B, II – B, III - B, IV – A or C I. Friction charging causes electrons to move from the animal fur to Object A, which becomes negative. II. Conduction charging causes excess electrons in the negatively charged object to be shared with Object A, so object A becomes negatively charged. III. Induction charging causes Object A to be oppositely charge, so using a positive object induces a negative charge on Object A. IV. Object A can be attracted to a negatively charged object if it oppositely charged positive, but also may be attracted by polarization if it is neutral. resistance (much like a drinking straw has less resistance than a narrow stirring straw.) 2. answer: A 9. answer: C Ordinarily only electrons can be moved from one body to another (movement of protons requires high-energy collisions in accelerators!). So a glass rod acquires a positive charge, when rubbed with silk, by losing electrons. DV = 8. answer: D Using the definition of resistance as the ratio of voltage divided by current: DV DV 0.40 volt R= Þ I= = = 0.004 A I R 100 W Converting amps to milliamps: 1 mA 0.004 A ´ -3 = 4.0 mA 10 A 4. answer: B By definition, electric field lines point towards negative charges and away from positive charges (because a test charge is always assumed to be positive). So sphere A must be negative and sphere B must be positive to create the field lines shown in the drawing. work 2 ´ 10 -4 work = 2 ´ 10 -3 J E= kq (9 ´ 10 9 )(q) 3600 = q = 4 ´ 10 -9 C = 4 nC 2 2 r 0.1 11. answer: D The electric field inside this parallel plate capacitor is constant (except at the left and right ends). The field points down because a positive test charge is used to define the direction of an electric field, and the positive test charge would be repelled by the positive charges above, and attracted by the negative charges below it. 12. answer: C DV = I(R1 + R2 ) 12 = 2(3 + R2 ) R2 = 3 W 13. answer: D PE = P(t) = IDV(t) = 10(120)(30) = 3.6 ´ 10 4 J 14. answer: B 5. answer: C Fe = 10 = 10. answer: C 3. answer: D The positively charged rod near the electroscope will attract some electrons into the knob (top) of the electroscope, leaving the electroscope polarized, but still neutral. The correct representation has negative signs in the knob, and positive signs in the leaves (bottom). PE work = q q -19 2 kq1q2 9 ´ 10 (1.6 ´ 10 ) = r2 (1.00 ´ 10 -8 )2 9 1 1 1 1 1 1 = + + = + Req R1 R2 R3 20 30 Fe = 2.30 ´ 10 -12 N 15. answer: B 6. answer: A I= number of electrons = charge fundamental charge DV 12 = = 0.60 A R 20 16. answer: A DV 2 12 2 = = 4.8 W R 30 q -8 ´ 10 -6 n= = = 5 ´ 1013 -19 e -1.6 ´ 10 P= 7. answer: B 17. answer: C Electrical resistance depends directly on temperature, length, and resistivity. It depends inversely on cross-sectional area. That is, the wider the diameter of resistor the less the Req = 12 W The ammeter must be placed in series with resistor #1 within the branch, so at position 3. The voltmeter must be placed in parallel with resistor #2, so at position 1. 18. answer: D 21a. circuit A: Req = 2 + 2 = 4 W -1 DV1 = I1R1 -1 -1 120 = 0.60R1 circuit B: Req = (2 + 2 ) = 1 W 21b. circuit C: Req = 1+ 1 = 2 W DV2 = I 2 R2 -1 -1 -1 circuit D: Req = (1 + 1 ) = 1 2 W 120 = I 2 (240) I 2 = 0.5 A 21c. P3 = I 3DV3 smallest resistance is circuit D 80 = I 3 (120) I 3 = 0.67 A 21d. 19a. Fe = R1 = 200 W kq1q2 (9 ´10 )(8 ´1.6 ´10 = r2 (5.9 ´10 -15 )2 9 -19 2 ) P = I1DV1 = 0.6(120) = 72 W = 424 N 19b. R1 R 200 I 0.6 V 120 P 72 kq1q2 (9 ´ 10 9 )(-1.6 ´ 10 -19 )(8 ´ 1.6 ´ 10 -19 ) = r2 (2.4 ´ 10 -14 )2 Fe = -3.2 N (to the right) R2 240 0.5 120 60 R3 180 0.67 120 80 Req 67.9 1.767 120 212 Fe = -19 kq1q3 (9 ´ 10 )(-1.6 ´ 10 )(8 ´ 1.6 ´ 10 = r2 (2.4 ´ 10 -14 + 5.9 ´ 10 -15 )2 Fe = -2.06 N (to the right) 9 Fe = -19 ) 22a. SF = 3.2 + 2.06 = 5.26 N (to the right) 19c. E= Fe 5.26 = = 3.29 ´ 1019 N/C, to the left -19 q0 (1.6 ´ 10 ) I2 = 0.5 A Req = RABC + RD = 8 +12 = 20 W 22c,d,e. 12 V Series: 20b. DV = DV1 + DV2 12 = 10 + DV2 DV2 = 2 v 20c. I = I1 = I 2 = 0.5 A DV1 = I1 R1 10 = 0.5R1 12 = 0.5Req Req = 24 W 20e. P2 = I 2 DV2 = 0.5(10) = 5 W table method: R R1 4 RB R 12 I 0.1 V 1.2 P 0.12 RC 12 0.1 1.2 0.12 RBC 24 0.1 2.4 0.24 RBC R 24 I 0.1 V 2.4 P 0.24 RD 12 0.2 2.4 0.48 RBCD 8 0.3 2.4 0.72 I 0.3 V 2.4 P 0.72 Parallel: R1 = 20 W 20d. DV = IReq RBC = RB + RC = 12 +12 = 24 W RABC = (RBC -1 + RA -1 )-1 = (24 -1 +12 -1 )-1 = 8 W 20a. ΔV1 = 10 v 22b. Series: I 0.5 V 2 P 1 RBCD R 8 RA 12 0.3 3.6 1.08 Req 20 0.3 6 1.80 R2 20 0.5 10 5 Req 24 0.5 12 6 Series: 23a. R12 = R1 + R2 = 6 + 4 = 10 W R123 = (R12 + R3 ) = (10 + 15 ) = 6 W R1 R 6 Req = R123 + R4 + R5 = 6 + 4 + 2 = 12 W R2 4 1.8 7.2 12.96 R12 10 1.8 18 32.4 R12 R 10 I 1.8 V 18 P 32.4 R3 15 1.2 18 21.6 R123 6 3 18 54 I 3 V 18 P 54 -1 -1 -1 -1 -1 -1 DV 36 I= = = 3A Req 12 P= I 1.8 V 10.8 P 19.44 Parallel: DV 2 36 2 = = 108 W Req 12 23b. I = I4 = I5 = 3 A Series: DV5 = I 5 R5 = 3(2) = 6 v P5 = I 5 DV5 = 3(6) = 18 W R123 R 6 DV4 = I 4 R4 = 3(4) = 12 v R4 4 3 12 36 R5 2 3 6 18 Req 12 3 36 108 P4 = I 4 DV4 = 3(12) = 36 W DV3 = DV123 = DV - DV45 = 36 - 18 = 18 v DV3 = I 3 R3 18 = I 3 (15) I 3 = 1.2 A 24a. P3 = I 3 DV3 = 1.2(18) = 21.6 W I1 = I 2 = I - I 3 = 3 - 1.2 = 1.8 A DV2 = I 2 R2 = 1.8(4) = 7.2 v P2 = I 2 DV2 = 1.8(7.2) = 12.96 W DV1 = I1 R1 = 1.8(6) = 10.8 v P1 = I1DV1 = 1.8(10.8) = 19.44 W 24b. q1 = +3 nC ´ 10 -9 C = +3 ´ 10 -9 C 1 nC Then complete the third table for the final series circuit. Only the known values are shown in bold. All others are found by multiplication or division. q2 = –1 nC ´ 10 -9 C = –1 ´ 10 -9 C 1 nC Then complete the second table. Notice the bottom row is simply taken from the table below it. r12 Using the table method, work through each resistor column from the first table, to the second, then to the third. Finally, complete the first table. Again, notice the bottom row is simply taken from the table below it E1 = kq1 E2 = kq2 r12 = = 9.0 ´ 10 9 (3 ´ 10 -9 ) = 108 N/C, right ( 0.5 m )2 9.0 ´ 10 9 (1´ 10 -9 ) ( 0.5 m )2 = 36 N/C, right ETOTAL = E1 + E2 = 108 N/C + 36 N/C = 144 N C, right 24c. k(3 nC)(2 nC) k(1 nC)(2 nC) = Þ r = 1.37 m (1+ r)2 (r)2