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MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 13: ELECTRIC DC CURRENT and RESISTANCE WORKSHEET SOLUTIONS 1. The heating coil of a hot water heater has a resistance of 50.0 ohm and operates at 120 V. What is its power rating? 1A. (1) P = V2 / R (2) P = (120 V)2 / 50.0 (3) P = 288 W 2. If a certain resistor obeys Ohm’s law, its resistance will change: (A) as voltage only across the resistor changes (B) as current only through the resistor changes (C) as both voltage and current change (D) as energy given off by the electrons in their collisions changes (E) none of the above 2A. (E) none of the above 3. A certain material is in a room at 27.0oC. If the absolute temperature K of the material is tripled, its resistance also triples. What is the value for , the temperature coefficient of resistivity? 3A. (1) R = Ro(1 + (T – To)) (2) (T – To) = (R/Ro) – 1 (3) R = 3Ro (4) (T – To) = (3Ro/Ro) – 1 (5) (T – To) = 2 (6) = 2 / (T – To) (7) = 2 / (900 K – 300 K) (8) = 2 / 600 K (9) = 0.0033 / oC 4. A platinum wire is utilized to determine the melting point of indium. The resistance of the platinum wire is 2.00 at 20.0oC and increases to 3.072 as the indium starts to melt. platinum = 3.92 x 10-3/oC. What is the melting temperature of indium (in oC)? 4A. (1) R = Ro(1 + (T – To)) (2) T = [(R – Ro) / (Ro)] + To (3) T = [(3.072 – 2.00 ) / (3.92x10-3/oC)(2.00 )] + 293 K (4) T = 136.7 K + 293 K (5) T = 429.7 K (6) Tc = T – 273 (7) Tc = 156.7oC 1|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 5. A color television set draws 3.15 A when connected to 220 V. What is the cost (with electrical energy at 7 cents/kWh) of running the color TV for 8 hours? 5A. (1) Determine joules used: P = IV (2) P = (3.15 A)(220 V) (3) P = 693 W = 693 J/s (4) (8 hours) (3600 sec / 1 hour) (693 J / 1 sec) = 2.00 x 107 J (5) 2.00 x 107 J (1 kWh) = 5.54 kWh -------------3.6 x 106 J (6) Determine cost: (5.54 kWh) (7 cents) (7) Cost = 38.8 cents 6. An electric car runs off a bank of 18.0 V batteries with total energy storage of 4.55 x 109 J. If the electric motor draws 6000 W in moving the car at a steady speed of 20 m/s, how far will the car go (in km) before it runs out of fuel? 6A. (1) Determine time car can move: Time = (4.55 x 107 J) (6000 W) (2) Time = (4.55 x 107 J) (1 sec / 6000 J) (3) Time = 7583 s (4) Determine distance car can move: (7583 s) (20 m/s) = 1.52 x 105 m (5) 1.52 x 105 m (1 km / 1000 m) = 152 km 7. How long does it take for 0.2 C to pass through a frictionless wire in a 9 amp circuit? 7A. (1) I = Q / t (2) t = Q / I (3) t = 0.2 C / 9A (4) t = 2.22 x 10-2 s 8. A metal wire has a radius of 0.45 mm, is 0.85 m long, and has a resistance of 15.0 ohms. What must be the length of a second wire made of the same type of metal if its radius is 0.10 mm and also has a resistance of 15.0 ohms? 8A. (1) R1 = L1 / A (2) = R1A / L1 (3) A = r2 (4) = R1r2 / L1 (5) = (15 ) (4.5 x 10-4 m)2 / 0.85 m (6) = 1.12 x 10-5 .m (7) R2 = L2 / A (8) L2 = R2A / 2|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 8A. (continued…) (9) L2 = R2(r2) / (10) L2 = (15 ) (1 x 10-4 m)2 / 1.12 x 10-5 .m (11) L2 = 4.20 x 10-2 m (12) L2 = 4.20 cm 3|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 13: ELECTRIC DC DIRECT CURRENT WORKSHEET SOLUTIONS 1. The following three appliances are connected to a 120 volt house circuit: (i) toaster, 1350 W (ii) coffee pot, 750 W (iii) microwave, 600 W If all were operated at the same time, what total current would they draw? 1A. (1) P = IV (2) I = P / V (3) I = 2850 W / 120 V (4) I = (2850 J/s) / (120 C/J) (5) I = 23.8 C/s = 23.8 A 2. A circuit contains a 10 volt battery, a 4 resistor, a 4.2 picofarad capacitor, a multimeter, and a switch, all in series. What will be the current reading 5 minutes after the switch is closed? 2A. zero 3. Three resistors, with values of 3.00, 6.00, and 12.0 ohms, respectively, are connected in parallel. What is the equivalent resistance of this combination? 3A. (1) 1/Rp = (1/Ri) (2) 1/Rp = 1 / (3 + 1 / (6 + 1 / (12 (3) 1/Rp = 7 / (12 (4) Rp = (12 / 7) (5) Rp = 1.7 Questions 4 - 5. Examine the circuit below. 4|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 4. Which resistor above is in series with resistor R? (A) R1 (B) R2 (C) R3 (D) R4 (E) R6 4A. R3 5. The resistors are tested, and are found to have the following values: R1 = 2 , R2 = 4 , R3 = 10 , R4 = 3 , R5 = 3 , R6 = 6 . What is the equivalent resistance for these resistors? 5A. RP (top part): 1 / RP = 1 / R1 + 1 / R2 (1) 1 / RP = 1 / (2 + 1 / (4 (2) 1 / RP = 2 / (4 ) + 1 / (4 ) (3) 1 / RP = 3 / (4 ) (4) RP = (4 /3) (5) RP = 1.33 RS (middle part): RS = R4 + R5 (6) RS = 3 + 3 (7) RS = 6 RP (bottom part): 1 / RP = 1 / RS + 1 / R6 (8) 1 / RP = 1 / (6 + 1 / (6 (9) 1 / RP = 2 / 6 (10) 1 / RP = 1 / 3 (11) RP = 3 RS (entire circuit): RS = R + RP + R3 + RP (12) RS = 2 + 1.33 + 10 + 3 (13) RS = 16.3 Questions 6 - 8. The circuit below is connected to a 20 volt battery. 6. How much is the single resistance representing all four of the original resistors? 6A. RS (top part): (1) RS = 3 + 9 (2) RS = 12 5|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 6A. (continued…) RP (middle part): (3) 1 / RP = 1 / (12 + 1 / (6 RP = 1 / (12 ) + 2 / (12 ) (5) 1 / RP = 3 / (12 ) (6) 1 / RP = 1 / (4 ) (7) RP = 4 RS (entire circuit): (8) RS = 6 + 4 RS = 10 7. The average current through the battery is: 7A. (1) V = I Req (2) I = V / Req (3) I = 20 V / 10 (4) I = 2 A 8. The total power dissipated by the circuit is: 8A. (1) P = I V (2) P = (2 A)(20 V) (3) P = 40 W VOLTAGE AND BATTERIES Voltage leaves the cathode of a battery at full voltage. Voltage returns to the anode of a battery at zero volts. Questions 9 - 14. A 20 volt battery is connected across a 5 ohm resistor, as shown. Questions 9 – 14. Diagram 6|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 9. Which end of the resistor is at the higher potential difference, A or B? 9A. Point A 10. The 5 ohm resistor consists of a block of material with terminals at A and B. The electric field inside this material must either be directed from A to B or from B to A. Which is it? 10A. From A to B. Electric field flows out of the positive end of the battery and into the negative end of the battery. 11. What is the direction of electrons through the resistor? 11A. It is directed from B to A through the resistor. 12. What is the voltage across the 5 ohm resistor? 12A. 20V. The battery supplies the voltage. 13. What is the current through the 5 ohm resistor? 13A. (1) V = I R (2) I = V / R (3) I = 20 V / 5 (4) I = 4 A 14. Electrons scatter off atoms in the resistor and transfer energy (heat) to them. What is rate of heat production, or power, dissipated in the 5 ohm resistor? 14A. (1) P = I V (2) P = (4 A)(20 V) (3) P = 80 W 7|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 14: MAGNETISM, INDUCTION, AND A/C CIRCUITS WORKSHEET SOLUTIONS 1. A circular loop carrying a current of 1.42 A is oriented in a magnetic field of 2.74 T. The loop has an area of 0.17 m2 and is mounted on an axis, perpendicular to the magnetic field which allows the loop to rotate. What is the torque on the loop when its plane is oriented at an 18o angle to the field? 1A. (1) = BIA(sin (2) = (2.74 T)(1.42 A)(0.17 m2)(sin(90-18))o (3) = 0.63 N.m 2. A current-carrying wire of length 0.41 m is positioned perpendicular to a uniform magnetic field. If the current is 12.5 A and it is determined that there is a resultant force of 4.05 N on the wire due to the interaction of the current and field, what is the magnetic field strength? 2A. (1) F = BIL(sin (2) F = BIL (3) B = F / IL (4) B = (4.05 N) / (12.5 A)(0.41 m) (5) B = 0.79 T 3. A charged particle, mass 3.51 x 10-25 kg and charge 3.24 x 10-19 C, moves in a circular orbit perpendicular to a uniform magnetic field of 0.75 T. Find the time for the proton to make one complete circular orbit. 3A. (1) T = (2m) / (qB) (2) T = (2)(3.51 x 10-25 kg) / (3.24 x 10-11 C)(0.75 T) (3) T = 9.08 x 10-6 s 4. An electron moves through a region of crossed electric and magnetic fields. The electric field E = 1225 V/m and is directed straight down. The magnetic field B = 1.15 T and is directed to the left. For what velocity v of the electron into the paper (perpendicular to the plane) will the electric force exactly cancel the magnetic force? 4A. (1) FE – FB = 0 (2) FE = FB (3) FE = Eq (4) FB = qvB(sin (5) FB = qvB (6) Eq = qvB v = E / B (8) v = (1225 V/m) / (1.15 T) (9) v = 1065 m/s 8|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 5. A proton is released such that its initial velocity is from left to right across this page. The proton’s path, however, is deflected in a direction toward the bottom edge of the page due to the presence of a uniform magnetic field. What is the direction of this field? Question 5. Diagram 5A. (B) into the page 6. Two parallel cables of a high voltage transmission line carry equal and opposite currents of 2215 A. The distance between the cables is 1.17 m. What is the magnetic force acting on a 21 meter length of each cable? 6A. (1) F / L = oI1I2 / 2r (2) F = oI1I2L / 2r (3) F = (4 x 10-7 T.m/A)(2215 A)(2215 A)(21 m) / (2) (1.17m) (4) F = 17.6 N 7. A 15 cm x 25 cm rectangular loop consists of 80 closely-wrapped turns which each carry a current of 0.75 A. The loop is oriented 20o parallel to a uniform magnetic field of 0.35 T directed in the positive y direction. What is the magnitude of the torque on the loop? 7A. (1) = BnIA(sin (2) A = l x w (3) A = (0.15 m)(0.25 m) (4) A = 3.75 x 10-2 m2 (5) = (0.35 T)(80)(0.75 A)(3.75 x 10-2 m2)(sin (90o – 20o) (6) = (0.35 T)(80)(0.75 A)(3.75 x 10-2 m2) (sin 70o) (7) = 0.74 N.m 9|Page PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 8. A superconducting solenoid is to be designed to generate a magnetic field of 14.7 T. If the solenoid winding has 1715 turns/meter, what is the required current? 8A. (1) B = onI (2) I = B / on (3) I = (14.7 T) / (4 x 10-7 A.m/T)(1715) (4) I = 6821 A 9. A positive charge and a negative charge enter into the page. The magnetic field is uniform and points to the left of the page. Describe the deflections of each charge. Example 9. Diagram 9A. The positive charge deflects toward the top of the page. The negative charge deflects towards the bottom of the page. While in the field, their paths are circular arcs. 10. The two charges now travel antiparallel to a uniform magnetic field, as shown below. Describe the deflections of each charge. Example 10. Diagram 10 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 10A. There is no deflection since the magnetic force is zero. The charges continue in straight line motion. Questions 11 - 12. A long straight 0.5 m wire passes north-south through a region where the 0.2 T magnetic field is uniform and directed out of the page. The 20 ampere current is directed to the bottom of the page. Example 11. Diagram 11. What is the direction of the magnetic force on the wire? 11A. To the left. 12. Calculate the magnitude of the force. 12A. (1) F = BIL(sin (2) F = BIL (3) F = (0.2 T)(20 A)(0.5 m) (4) F = 2 N 13. A square loop 1.50 m on a side is placed in a magnetic field of 0.150 T. If the field makes an angle of 12.0o with the normal to the plane of the loop, determine the magnetic flux through the loop. 13A. (1) = BA(cos (2) A = s2 (3) A = (1.50 m)2 (4) A = 2.25 m2 (5) = (0.150 T)(2.25 m2)(cos 12o) (6) = 3.30 x 10-1 T.m2 11 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 14. A 125-turn solenoid (r = 0.35 m) is placed in a magnetic field of 0.225 T. The field makes a 75o angle with the normal to the plane of the loop. The magnetic field then decreases to zero in 0.1 s. Determine the average voltage induced during this time interval. 14A. (1) = BA(cos (2) A = r2 (3) A = (0.35 m)2 (4) A = 0.385 m2 (5) = (B)Acos (6) Bo = 0.225 T (7) B = 0 T (8) B = 0.225 T – 0 T (9) B = 0.225 T (10) = (0.225 T)(0.385 m2)(cos 75o) (11) = 2.24 x 10-2 T.m2 (12) = - N / t (13) = (125)(2.24 x 10-2 T.m2) / (0.1 s) (14) = 28.0 V 15. A wire loop (r = 0.4 m) is placed in a magnetic field of 1.20 T. The field makes a 45o angle to the plane of the loop. The wire loop is shrunk to a radius of 0.05 m in 0.07 s. Determine the average voltage induced during this time interval. 15A. (1) = BA(cos (2) Ao = r2 (3) Ao = (0.4 m)2 (4) Ao = 5.03 x 10-1 m2 (5) A = r2 (6) A = (0.05 m)2 (7) A = 7.85 x 10-3 m2 (8) = B(A)cos (9) A = 5.03 x 10-1 m2 – 7.85 x 10-3 m2 (10) A = 4.95 x 10-1 m2 (11) = (1.20 T)(4.95 x 10-1 m2)(cos 45o) (12) = 4.20 x 10-1 T.m2 (13) = / t (14) = (4.20 x 10-1 T.m2) / (0.07 s) (15) = 6.00 V 16. A 0.41 m wire is moved perpendicular to a 0.58 T magnetic field at a speed of 2.50 m/s. What emf is induced across the ends of the wire? 16A. (1) = BLv (2) = (0.58 T)(0.41 m)(2.50 m/s) (3) = 0.59 V 12 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 17. A 95 Hz RC circuit flows through a 40 F capacitor. The maximum current through the circuit is 24 amperes. Calculate the average voltage. 17A. (1) XC = 2fC (2) XC = 2(95 Hz)(40 x 10-6 F) (3) XC = 2.39 x 10-2 (4) V = I XC (5) IRMS = IMAX / (2)1/2 (6) IRMS = 24 A / (2)1/2 (7) IRMS = 17.0 A (8) V = (17.0 A)(2.39 x 10-2 ) (9) V = 0.41 volts 18. A 105 Hz RL circuit flows through a 35 mH inductor. Calculate the circuit’s inductive reactance. 18A. (1) XL = 2fL (2) XL = 2(105 Hz)(35 x 10-3 H) (3) XL = 23.1 19. A 105 Hz RLC circuit contains a 10 resistor. The current flows through a 32 F capacitor and a 41 mH inductor. Calculate the circuit’s impedance. 19A. (1) XL = 2fL (2) XL = 2(105 Hz)(41 x 10-3 H) (3) XL = 27.0 (4) XC = 2fC (5) XC = 2(105 Hz)(32 x 10-6 F) (6) XC = 2.11 x 10-2 (7) R = 10 (given in problem) (8) Z = [ R2 + (XL - XC)2 ]1/2 (9) Z = [ (10 )2 + (27.0 – 2.11 x 10-2 )2 ]1/2 (10) Z = 28.8 13 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTERS 13 - 14: ELECTRIC D/C, MAGNETISM, AND ELECTRIC A/C QUIZ SOLUTIONS 1. A color television set draws about 3.25 A when connected to 120 V. What is the cost (with electrical energy at 8 cents/kWh) of running the color TV for 14 hours? 1A. (1) Determine joules used: P = IV (2) P = (3.25 A)(120 V) (3) P = 390 W = 390 J/s (4) (14 hours) (3600 sec / 1 hour) (390 J / 1 sec) = 1.97 x 107 J (5) 1.97 x 107 J (1 kWh) -------------- = 5.47 kWh 3.6 x 106 J (6) Determine cost: (5.47 kWh) (8 cents) (7) Cost = 43.7 cents 2. An electric car is designed to run off a bank of 18.0 V batteries with total energy storage of 3.15 x 107 J. If the electric motor draws 9200 Watts in moving the car at a steady speed of 22 m/s, how far will the car go (in km) before it runs out of energy? 2A. (1) Determine time car can move: Time = (3.15 x 107 J) (9200 W) (2) Time = (3.15 x 107 J) (1 sec / 9200 J) (3) Time = 3424 s (4) Determine distance car can move: (3424 s) (22 m/s) = 7.53 x 104 m (5) 7.53 x 104 m (1 km / 1000 m) = 75.3 km 3. The heating coil of a hot water heater has a resistance of 27.0 ohm and operates at 180 V. What is its power rating? 3A. (1) P = V2 / R (2) P = (180 V)2 / 27.0 (3) P = 1200 W 4. Three resistors, with values of 4.00, 6.00, and 18.0 ohms, respectively, are connected in parallel. What is the overall resistance of this combination? 4A. (1) 1 / Req = 1 / R1 + 1 / R2 + 1 / R3 (2) 1 / Req = 1 / (4 + 1 / (6 + 1 / (18 (3) 1 / Req = 9 / (36 (4) 1 / Req = 17 / (36 (5) Req = (36/ 17) 14 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL Questions 5 - 6. A circuit obeys Ohm’s law. It contains a 12 volt battery and a 4 resistor. 5. The current in this circuit is: 5A. (1) V = IR (2) V = IReq (3) I = V / Req (4) I = 12V / 4 (5) I = 3.0 A 6. The total power dissipated by the circuit is: 6A. (1) P = IV (2) P = (3A)(12V) (3) P = 36 W 7. The following three appliances are connected to a 120 volt house circuit: (i) toaster, 1325 W (ii) coffee pot, 950 W; and (iii) microwave, 675 W. If all were operated at the same time, what total current would they draw? 7A. (1) P = IV (2) I = P / V (3) I = 2950 W / 120 V (4) I = (2950 J/s) / (120 C/J) (5) I = 24.58 C/s = 24.6 A 8. How long does it take for 1.5 C to pass through a frictionless wire in a 12 amp circuit? 8A. (1) I = Q / t (2) t = Q / I (3) t = 1.5 C / 12 A (4) t = 0.125 s 9. A circular loop carrying a current of 3.25 A is oriented in a magnetic field of 1.50 T. The loop has an area of 0.17 m2 and is mounted on an axis, perpendicular to the magnetic field, which allows the loop to rotate. What is the torque on the loop when its plane is oriented at a 42.0o angle to the field? 9A. (1) = BIA(sin (2) = (1.50 T)(3.25 A)(0.17 m2)(sin(90- 42))o (3) = (1.50 T)(3.25 A)(0.17 m2)(sin(48))o (4) = 0.62 N.m 15 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 10. A current-carrying wire of length 0.472 m is positioned perpendicular to a uniform magnetic field. If the current is 12.75 A and it is determined that there is a resultant force of 7.21 N on the wire due to the interaction of the current and field, what is the magnetic field strength? 10A. (1) F = BIL(sin) (2) B = F / IL (3) B = (7.21 N) / (12.75 A)(0.472 m) (4) B = 1.20 T 11. A proton, mass 1.67 x 10-27 kg and charge 1.60 x 10-19 C, moves in a circular orbit perpendicular to a uniform magnetic field of 3.15 T. Find the time for the proton to make one complete circular orbit. 11A. (1) T = (2m) / (qB) (2) T = (2)(1.67 x 10-27 kg) / (1.60 x 10-19 C) (3.15T) (3) T = 2.08 x 10-8 s (4) T = 20.8 ns 12. An electron moves through a region of crossed electric and magnetic fields. The electric field E = 1375 V/m and is directed straight down. The magnetic field B = 0.72 T and is directed to the left. For what velocity v of the electron into the paper (perpendicular to the plane) will the electric force exactly cancel the magnetic force? 12A. (1) FE – FB = 0 (2) FE = FB (3) FE = Eq (4) FB = qvB(sin (5) FB = qvB (6) Eq = qvB v = E / B (8) v = (1375 V/m) / (0.72 T) (9) v = 1910 m/s 13. Two parallel cables of a high voltage transmission line carry equal and opposite currents of 2300 A. The distance between the cables is 2.15 m. What is the magnetic force acting on a 47.0 meter length of each cable? 13A. (1) F / L = oI1I2 / 2r (2) F = oI1I2L / 2r (3) F = (4 x 10-7 T.m/A)(2300 A)(2300 A)(47.0 m) / (2)(2.15 m) (4) F = 23.1 N 16 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 14. A square loop 1.35 m on a side is placed in a magnetic field of 0.250 T. If the field makes an angle of 18.0o with the normal to the plane of the loop, determine the magnetic flux through the loop. 14A. (1) = BA(cos (2) A = s2 (3) A = (1.35 m)2 (4) A = 1.83 m2 (5) = (0.250 T)(1.83 m2)(cos 18o) (6) = 4.33 x 10-1 T.m2 15. A wire loop (r = 0.5 m) is placed in a magnetic field of 1.45 T. The field makes a 45o angle to the plane of the loop. The wire loop is shrunk to a radius of 0.08 m in 0.07 s. Determine the average voltage induced during this time interval. 15A. (1) = BA(cos (2) Ao = r2 (3) Ao = (0.5 m)2 (4) Ao = 7.85 x 10-1 m2 (5) A = r2 (6) A = (0.08 m)2 (7) A = 2.01 x 10-2 m2 (8) = B(A)cos (9) A = 7.85 x 10-1 m2 – 2.01 x 10-2 m2 (10) A = 7.65 x 10-1 m2 (11) = (1.45 T)(7.65 x 10-1 m2)(cos 45o) (12) = 7.85 x 10-1 T.m2 (13) = - N/t (14) = (1)(7.85 x 10-1 T.m2) / (0.07 s) (15) = 11.2 V 16. A 0.52 m wire is moved perpendicular to a 1.72 T magnetic field at a speed of 3.15 m/s. What emf is induced across the ends of the wire? 16A. (1) = BLv (2) = (1.72 T)(0.52 m)(3.15 m/s) (3) = 2.82 V 17. A 115 Hz RC circuit flows through a 30 F capacitor. The maximum current through the circuit is 18 amperes. Calculate the average voltage. 17A. (1) XC = 2fC (2) XC = 2(115 Hz)(30 x 10-6 F) (3) XC = 2.17 x 10-2 (4) V = I XC 17 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL 17A. (continued…) (5) IRMS = IMAX / (2)1/2 (6) IRMS = 18 A / (2)1/2 (7) IRMS = 12.7 A (8) V = (12.7 A)(2.17 x 10-2 ) (9) V = 0.28 volts 18. An 85 Hz RL circuit flows through a 55 mH inductor. The average current through the circuit is 22 amperes. Calculate the circuit’s maximum voltage. 18A. (1) XL = 2fL (2) XL = 2(85 Hz)(55 x 10-3 H) (3) XL = 29.4 (4) V = I XL (5) V = (22 A)(29.4 ) (6) V = 646.2 V (7) VRMS = VMAX / (2)1/2 (8) VMAX = V (2)1/2 (9) VMAX = (646.2 V)(2)1/2 (10) VMAX = 913.9 volts 19. A 125 Hz RLC circuit contains: a 25 resistor, a 22 F capacitor, and a 30 mH inductor. The maximum current is 25 amperes. Calculate the circuit’s average voltage. 19A. (1) XL = 2fL (2) XL = 2(125 Hz)(30 x 10-3 H) (3) XL = 23.6 (4) XC = 2fC (5) XC = 2(125 Hz)(22 x 10-6 F) (6) XC = 1.73 x 10-2 (7) R = 25 (given in problem) (8) Z = [ R2 + (XL - XC)2 ]1/2 (9) Z = [ (25 )2 + (23.6 – 1.73 x 10-2 )2 ]1/2 (10) Z = 33.2 IRMS = IMAX / (2)1/2 (12) IRMS = (25 A) / (2)1/2 (13) IRMS = 17.7 A (14) V = I Z (15) V = (17.7 A)(33.2 ) (16) V = 586.8 volts 18 | P a g e PHYSICS
