Circuits Practice Questions Review Formula’s RB BL RC AB RP V IR P IV I R I rR E Pt ...
Download ReportTranscript Circuits Practice Questions Review Formula’s RB BL RC AB RP V IR P IV I R I rR E Pt ...
Circuits Practice Questions Review Formula’s RB BL RC AB 1 RP V IR P IV I R 2 I rR E Pt 1 1 ... R1 Rn R s R1 ... R n V Charging 2 R 1 CS 1 1 ... C1 t Q t Q 0 1 e RC t Q t Q 0 e RC Cn C P C 1 ... C n Q CV U 1Q I t I 0e 2 2 C 1 2 QV 1 2 CV 2 t RC Discharging Question A wire made of brass and another wire made of silver have the same length, but the diameter of the brass wire is 4 times the diameter of the silver wire. The resistivity of brass is 5 times greater than the resistivity of . Silver. If RB denotes the resistance of the brass wire and RS denotes the resistance of the silver wire, which of the following is true? a) b) c) d) e) RB=5/16 RS RB=4/5 RS RB=5/4 RS RB=5/2 RS RB=16/5 RS Let ρs denote the resistivity of silver and let As denote the cross-sectional area of the silver wire. RB BL AB 5s L 2 4 As 5 16 Question For a ohmic conductor, doubling the voltage without changing the resistance will cause the current to? a) b) c) d) e) Decrease by a factor of 4 Decrease by a factor of 2 Remain unchanged Increase by a factor of 2 Increase by a factor of 4 I V R Therefore doubling the voltage, doubles the current. Question If a 60 watt light bulb operates at a voltage of 120V, what is the resistance of the bulb? a) b) c) d) e) 2Ω 30Ω 240Ω 720Ω 7200Ω P V 2 R R V P 120V 2 2 60W 240 Question A battery whose emf is 40V has an internal resistance of 5Ω. If this battery is connect to a 15Ω resistor R, what will the voltage drop across R be? a) b) c) d) e) 10V 30V 40V 50V 70V I I V IR rR 40V 5 15 2A V IR 2 A 1 5 3 0V Question Three resistors are connected to a 10-V battery as shown below. What is the current through the 2.0 Ω resistor? 4.0Ω a) 0.25A b) 0.50A R s R1 ... R n 4.0Ω c) 1.0A V IR d) 2.0A ε=10V 2.0Ω e) 4.0A Since all resistors are in series, the amount of current that passes through any one of them is the same. So we need to simply the circuit to determine that current. R s R1 ... R n 4 4 2 V IR I 10 V R 1 0V 10 1A Question Determine the equivalent resistance between points a and b? a) b) c) d) e) 0.167Ω 0.25 Ω 0.333 Ω 1.5 Ω 2Ω R1 1 1 12 3 1 4 R 2 3 3 6 R3 12Ω 1 1 6 3Ω 4Ω 3Ω 1 3 2 1 RP 1 R1 1 R2 Question Three identical light bulbs are connected to a source of emf, as shown in the diagram above. What will happen if the middle bulb burns out? a) All the bulbs will go out b) The light intensity of the other two bulbs will decrease (but they won’t go out). c) The light intensity of the other two bulbs will increase. d) The light intensity of the other two bulbs will remain the same. e) More current will be drawn from the source emf. If each bulb has a resistance of R, then each individual bulb will draw ε/R. This will be unchanged if any individual bulb goes out. (less current will be drawn from the battery, but the same amount of current will pass through each bulb) Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected. a) Is Higher b) Is Lower c) Is The Same d) Don’t know Bulbs in parallel are like resistors in parallel. Therefore since the total resistance of parallel resistors is lower, and the voltage ins the same, then the current must increase (double). Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the current from the battery compared to when only one bulb was connected. a) Is Higher b) Is Lower c) Is The Same d) Don’t know Since the total resistance goes up by two, the current must go down by two. Therefore lower. Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in parallel to the first light bulb. After the second light bulb is connected, the power output from the battery (compared to when only one bulb was connected) a) Is four times higher b) Is twice as high c) Is the same d) Is half as much e) Is one quarter as much f) Don’t know Since the current goes up by two. Therefore the Power goes up by two P IV I R 2 V 2 R Question An ideal battery is hooked to a light bulb with wires. A second identical light bulb is connected in series with the first light bulb. After the second light bulb is connected, the light from the first bulb (compared to when only one bulb was connected) a) is four times as bright b) is twice as bright c) is the same d) is half as bright e) is one quarter as bright P IV I R 2 V 2 R Since the voltage goes down by a factor of two, the power goes down by 4. Therefore dimmer. Question What is the voltage drop across the 12 ohm resistor in the portion of the circuit shown? a) b) c) d) e) 24V 36V 48V 72V 144V R1 1 8 1 4 1 8 R2 2 2 4 8Ω 4Ω 2Ω 12A 1 8Ω 2 1 RP 1 R1 1 R2 V IR Since the top branch has 4Ω of resistance and the bottom branch has 12Ω of resistance, therefore 3 times as much current will flow through the 4Ω resistor than the 12Ω resistor. This breaks down the 12A into 9A up and 3A down. So form V=IR we have V=(3A)(12Ω)=36V 12Ω We could also have produced a RT and found 3Ω value, therefore a voltage drop of (3Ω)(12A)=36V on both the upper and lower branch Question What is the current through the 8Ω resistor in the circuit shown? a) b) c) d) e) 0.5A 1.0A 1.25A 1.5A 3.0A 2Ω V IR 2Ω 2Ω 2Ω 2Ω 2Ω 8Ω a 2Ω 24V 2Ω b Since points a and b are grounded, they are all at the same potential (call it zero). Travelling from b to a across the battery , the potential increases by 24V, therefore it must decrease by 24V across the 8Ω resistor as we reach point a. Therefore I=V/R =(24V)/(8Ω)=3A. Question How much energy is dissipated as heat in 20 seconds by a 100 Ω resistor that carries a current of 0.5A? a) b) c) d) e) 50 J 100 J 250 J 500 J 1000J E Pt P I R 2 P I R 2 0.5 A 2 100 2 5W 25 J s E Pt J 25 20 s s 500 J Question What is the time constant for the circuit shown? a) b) c) d) e) 0.01 s 0.025 s 0.04 s 0.05 s 0.1 s RT 50 200 250 RC 250 2 10 0.05 s 200Ω 50Ω 200 uF RC 50V 4 F Question A 100Ω, 120Ω, and 150Ω resistor are connected to a 9-V battery as in the circuit shown below. Which of the three resistors dissipates the most power? 2 We can also answer2 thisVquestion faster P IV I R by noticing that the voltage R drops across a) The 100Ω resistor the 100Ω and then across the parallel b) The 120Ω resistor 2 2 2 V V100Ω has a higher V resistors. Since the c) The 150Ω resistor P1 P2 P3 R R value than the parallel combination, R it will d) Both the 120Ω and the 150Ω 2 2 2 have a5 .4 larger V voltage V than the e) All dissipate the same power 3 .6drop 3 .6V P=IV, it dissipates combination so via the 1 0 0 1 2 0 100Ω 150 most power. 0 .2 9W 120Ω 9V 0 .1 0 8W 0 .0 8 6W 150Ω Therefore the 100Ω resistor dissipates the most amount of power. Question A 1.0 F capacitor is connected to a 12 V power supply until it is fully charged. The capacitor is then disconnected from the power supply, and then is used to power a toy car. The average drag force on the car is 2 N. about how far will the car go? a) b) c) d) e) 36m 72m 144m 24m 12m First we find the energy stored in the capacitor UC 1 CV 2 2 W Fd UC 1 2 1 CV This energy is the Work used in moving the car a fixed distance. 2 2 1 .0 F 72 J 1 2V 2 W Fd 72 J 2 N d d 36 m Question Three capacitors are connected to a 9 V power supply as shown. How much charge is stored by this system of capacitors? a) b) c) d) e) 3 uC 30 uC 2.7 uC 27 uC 10 uC Simplify the circuit to find equivalent capacitance. 1 1 CS ... C1 1 Cn C P C 1 ... C n Q CV C1=2uF 1 C EQ C1 C 2 CT 2 F 4 F C EQ 6 F 1 1 6 F 1 C3 1 6 F CT 3 F 9V C2=4uF To determine the charge , we use Q=CV Q CV 3 F C3=6uF 27 C 9V Question What is the resistance of an ideal ammeter and an ideal voltmeter? Measures current Ideal Ammeter a) b) c) d) e) zero infinite zero infinite 1Ω Ideal Voltmeter a) b) c) d) e) infinite zero zero infinite 1Ω Measures Voltage An ammeter is placed in series with other circuit components. In order for the ammeter not to itself resist current and change the total current in the circuit, you want it to have zero resistance. A voltmeter is placed in parallel with other circuit components. If it had a low resistance, the current will flow through it instead of the other circuit element. So you want it to have infinite resistance to it won’t affect the circuit element being measured. Question A light bulb is rated at 100 W in Canada, where the standard wall outlet voltage is 120V. If this bulb was plugged in in england, where standard wall outlet voltage is 240V, which of the following would be true? a) b) c) d) e) The bulb would be ¼ as bright. The bulb would be ½ as bright. The bulb’s brightness would the same. The bulb would be trice as bright. The bulb would be 4 times as bright P IV I R 2 V 2 R Your first instinct is to say that because brightness depends on power, the bulb is exactly as bright. But that is not correct. The power of the bulb can change. The resistance of a light bulb is a property of the bulb itself, and so will not change no matter what the bulb is hooked to. Question A current of 6.4A flows in a segment of copper wire. The number of electrons crossing a cross-sectional area of the wire every second is about? a) b) c) d) e) 6.4 4 x 1019 4 x 10-19 6.4 x 1019 6.4 x 10-19 I q t Since each electron carries a charge of 1.6 x 10-19 C. The we calculate that I 6 .4 A q t N 1 .6 1 0 1s N 4 10 19 19 C Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors a) What is the resistance of the toaster and blender at their rated voltages? b) Determine the current that is achieved in the blender when it is working properly. c) Create a circuit that will make both devices work simultaneously. d) What power must the battery supply to run your circuit? Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors a) What is the resistance of the toaster and blender at their rated voltages? Since we know the Power requirements , let’s us a power formula: P V 2 R 2 R to a ster 2 V T o a ster R b len d er PT o a ster 1 2 0V 1 2 0 0W 12 2 V b len d er Pb len d er 6 0V 1 2 0W 30 2 Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors b) Determine the current that is achieved in the blender when it is working properly. This is a job for Ohm’s Law: V IR I V R 6 0V 30 2A Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors c) Create a circuit that will make both devices work simultaneously. Required to give a current of 2A for blender 60Ω 60Ω Toaster ε=120V blender Question In a house trailer there is a 120V battery available for use. A toaster is designed to work properly as 120V, where it is rated at 1200W, and a 120W blender which is only designed to work properly at 60V. You look around the trailer and find a supply of 60Ω resistors d) What power must the battery supply to run your circuit? From Ohm’s Law, we can determine current in toaster V IR I V R Since the blender had a current of 2A, this gives the circuit a total current of 12A 1 2 0V 12 10 A Using P=IV, we have: P IV 12 A 12 0V 1440W Question Given the following circuit 10Ω 10Ω a 20Ω 40Ω ε=120V a) b) c) d) e) 100Ω b 10Ω At what rate does the battery deliver energy to the circuit? Determine the current through the 20 Ω resistor. i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher? Determine the energy dissipated by the 100 Ω resistor in 10 s Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is 0. 45 Ωm, determine its radius. Question Given the following circuit a) At what rate does the battery deliver energy to the circuit? Recall: Rate is Power 10Ω P IV 10Ω We need to write this as a simple circuit with a RT so that we can determine the current, I by using V=IR. 20Ω 40Ω ε=120V 100Ω RT 10Ω R T 10 10 10 1 1 40 60 1 20 100 P IV 2 A 120V 240W I V R 120V 60 2A Question Given the following circuit b) Determine the current through the 20 Ω resistor. Recall: from Question a) we have a 2A current The ratio of the resistance of left side to right side is 40:120 or 1:3 10Ω We can determine the voltage drop . 2A 10Ω 20Ω 40Ω ε=120V 100Ω 2A 10Ω 120V(2A)(10Ω)(2A)(10Ω)(2A)(10Ω) =60V Then we use V=IR to find each individual current. Therefore ¼ of 2 A goes through the right and ¾ of 2A goes through the left. Thus (1/4)(2A)=0.5A passes through the 20Ω resistor. OR Question Given the following circuit c) i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher? 10Ω a 10Ω 20Ω 40Ω ε=120V 100Ω 10Ω i) V a V b IR 20 IR100 0.5 A 20 0.5 A 1 00 6 0V b ii) Point a is at a higher potential. Since current flows from a high potential to a low potential. Question Given the following circuit d) Determine the energy dissipated by the 100 Ω resistor in 10 s 10Ω Recall: Energy equals Power multiplied by Time E Pt a 10Ω 20Ω 40Ω ε=120V 100Ω P I R 2 10Ω E I Rt 2 0.5 A 250 J 2 100 10 s b Question Given the following circuit e) Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is 0. 45 Ωm, determine its radius. 10Ω a RB BL A r AB 2 10Ω 20Ω 40Ω ε=120V RB r 100Ω BL r BL RB 10Ω 2 r 0.45 m 0.04 m 1 00 3 7.6 10 m b Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. S 10Ω a b r C ε R a) Determine the current through r at time t=0. b) Compute the time required for the charge on the capacitor to reach one-half its final value. c) When the capacitor is fully charged, which plate is positive? d) Determine the electric potential energy stored in the capacitor when the current r is zero. When the current through r is zero, the switch S is moved to b [t=0]. e) Determine the current through R as a function of time. f) Find the power dissipated in R as a function of time. Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. a) Determine the current through r at time t=0. S 10Ω V IR a b r C ε I r 10 R Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. b) Compute the time required for the charge on the capacitor to reach one-half its final value. S 10Ω t rC Q t Qi 1 e a b r C Therefore we need t r 10 C 1 e e ε 1 2 t r 10 C 1 2 1 ln r 10 C 2 t t r 10 C ln 2 R Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. c) When the capacitor is fully charged, which plate is positive? S 10Ω a b r + C ε R Because the top plate is connected to the positive end of the battery, the top is positive. Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. d) Determine the electric potential energy stored in the capacitor when the current r is zero. S 10Ω U 1 a b CV r 2 2 C R ε When the current through r is zero, the capacitor is fully charged, with voltage across the plates matching the emf of the battery. U 1 2 C 2 Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. When the current through r is zero, the switch S is moved to b [t=0]. e) Determine the current through R as a function of time. S 10Ω a b I t I 0e r t RC C R ε The current established by the discharging capacitor decreases exponentially. I t I 0e t RC e RC R t Question Given the following circuit with the switch S turned to point a at time t=0. Express all answers in terms of C, r, R, ε, and constants. When the current through r is zero, the switch S is moved to b [t=0]. f) Find the power dissipated in R as a function of time. S 10Ω a b P I R 2 r I t I 0e t C R ε RC P I t R 2 2 t e R R 2 R e 2t RC Question 200Ω 300Ω ε= 9V 400Ω 500Ω a) Simplify the above circuit so that it consists of one equivalent resistor and the battery. b) What is the total current through this circuit? c) Find the voltage across each resistor. d) Find the current through each resistor. e) The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same. Question 200Ω 300Ω ε= 9V REQ 400Ω 500Ω a) Simplify the above circuit so that it consists of one equivalent resistor and the battery. 1 1 R 400 R 2000 9 2 2 2 .2 1 1 500 R 1 200 R 120 1 300 R EQ 2000 120 9 R E Q 3 4 2 .2 Question 200Ω 300Ω ε= 9V 400Ω 500Ω b) What is the total current through this circuit? V IR I V R 9V 3 4 2 .2 0 .0 2 6 3 A REQ =342.2Ω Question 200Ω c) Find the voltage across each resistor. d) Find the current through each resistor. Let’s use a VIR chart 300Ω ε= 9V 500Ω V I R 200Ω 3.16V 0.0158A 200Ω 300Ω 3.16V 0.0105A 300Ω 400Ω 0.0146A 400Ω 500Ω 5.84V 5.84V 0.0117A 500Ω Total 9V 0.0263A 342.2Ω 1 R 1 200 R 120 1 300 V IR 0.0263 A 120 3.16V 400Ω V 9V 3.16V 5.84V Question 200Ω e) The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same. 300Ω ε= 9V 400Ω 500Ω By removing a resistor from a parallel set, we actually increase the resistance of the total circuit. Therefore by Ohm’s law if the voltage remains the same and the resistance increases, the total current must decrease Now through the 200Ω set, the total resistance remains the same, yet the current decreases, therefore the voltage across each resistor decreases as well as the current .