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Circuits Practice Questions

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Review Formula’s Charging Discharging

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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 R B denotes the resistance of the brass wire and R S denotes the resistance of the silver wire, which of the following is true? Question a)R B =5/16 R S b)R B =4/5 R S c)R B =5/4 R S d)R B =5/2 R S e)R B =16/5 R S Let ρ s denote the resistivity of silver and let A s denote the cross-sectional area of the silver wire.

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For a ohmic conductor, doubling the voltage without changing the resistance will cause the current to? Question a)Decrease by a factor of 4 b)Decrease by a factor of 2 c)Remain unchanged d)Increase by a factor of 2 e)Increase by a factor of 4 Therefore doubling the voltage, doubles the current.

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If a 60 watt light bulb operates at a voltage of 120V, what is the resistance of the bulb? Question a)2 Ω b)30 Ω c)240 Ω d)720 Ω e)7200 Ω

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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? Question a)10V b)30V c)40 V d)50V e)70V

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Question Three resistors are connected to a 10-V battery as shown below. What is the current through the 2.0 Ω resistor? a)0.25A b)0.50A c)1.0A d)2.0A e)4.0A 4.0Ω 2.0Ω ε=1 0V 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.

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Determine the equivalent resistance between points a and b? Question a)0.167Ω b)0.25 Ω c)0.333 Ω d)1.5 Ω e)2 Ω 12Ω 4Ω4Ω 3Ω3Ω 3Ω3Ω

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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? Question 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)

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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).

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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. Question 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.

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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) Question 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

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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) Question 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 Since the voltage goes down by a factor of two, the power goes down by 4. Therefore dimmer.

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What is the voltage drop across the 12 ohm resistor in the portion of the circuit shown? Question a)24V b)36V c)48V d)72V e)144V 8Ω8Ω 8Ω8Ω 2Ω2Ω 12Ω 4Ω4Ω 12A 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 We could also have produced a R T and found 3Ω value, therefore a voltage drop of (3Ω)(12A)=36V on both the upper and lower branch

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What is the current through the 8Ω resistor in the circuit shown? Question a)0.5A b)1.0A c)1.25A d)1.5A e)3.0A 2Ω2Ω 8Ω8Ω 2Ω2Ω 2Ω2Ω 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. 2Ω2Ω2Ω2Ω 2Ω2Ω 2Ω2Ω 2Ω2Ω 24V a b

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Question How much energy is dissipated as heat in 20 seconds by a 100 Ω resistor that carries a current of 0.5A? a)50 J b)100 J c)250 J d)500 J e)1000J

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What is the time constant for the circuit shown? Question a)0.01 s b)0.025 s c)0.04 s d)0.05 s e)0.1 s 200Ω 50V 200 uF 50Ω

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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? Question a)The 100Ω resistor b)The 120Ω resistor c)The 150Ω resistor d)Both the 120Ω and the 150Ω e)All dissipate the same power 100Ω 120Ω 9V 150Ω Therefore the 100Ω resistor dissipates the most amount of power. We can also answer this question faster by noticing that the voltage drops across the 100Ω and then across the parallel resistors. Since the 100Ω has a higher value than the parallel combination, it will have a larger voltage drop than the combination so via P=IV, it dissipates the most power.

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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)36m b)72m c)144m d)24m e)12m First we find the energy stored in the capacitor This energy is the Work used in moving the car a fixed distance.

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Question Three capacitors are connected to a 9 V power supply as shown. How much charge is stored by this system of capacitors? a)3 uC b)30 uC c)2.7 uC d)27 uC e)10 uC C 1 =2uF C 2 =4uF C 3 =6uF 9V Simplify the circuit to find equivalent capacitance. To determine the charge, we use Q=CV

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Question What is the resistance of an ideal ammeter and an ideal voltmeter? a)zero b)infinite c)zero d)infinite e)1Ω a)infinite b)zero c)zero d)infinite e)1Ω Ideal Ammeter Ideal Voltmeter Measures current 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.

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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)The bulb would be ¼ as bright. b)The bulb would be ½ as bright. c)The bulb’s brightness would the same. d)The bulb would be trice as bright. e)The bulb would be 4 times as bright 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.

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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)6.4 b)4 x c)4 x d)6.4 x e)6.4 x Since each electron carries a charge of 1.6 x C. The we calculate that

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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?

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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:

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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:

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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. 60Ω blender Toaster ε=1 20V 60Ω Required to give a current of 2A for blender

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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 Since the blender had a current of 2A, this gives the circuit a total current of 12A Using P=IV, we have:

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Question Given the following circuit a)At what rate does the battery deliver energy to the circuit? b)Determine the current through the 20 Ω resistor. c)i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher? d)Determine the energy dissipated by the 100 Ω resistor in 10 s e)Given that the 100 Ω resistor is a solid cylinder that’s 4 cm long, composed of a material whose resistivity is Ωm, determine its radius. 10Ω 20Ω 100Ω 40Ω 10Ω ε=1 20V a b

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Question Given the following circuit a)At what rate does the battery deliver energy to the circuit? 10Ω 20Ω 100Ω 40Ω 10Ω ε=1 20V Recall: Rate is Power We need to write this as a simple circuit with a R T so that we can determine the current, I by using V=IR. RTRT

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Question Given the following circuit b) Determine the current through the 20 Ω resistor. 10Ω 20Ω 100Ω 40Ω 10Ω ε=1 20V Recall: from Question a) we have a 2A current 2A The ratio of the resistance of left side to right side is 40:120 or 1:3 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. 120V- (2A)(10Ω)- (2A)(10Ω)- (2A)(10Ω) =60V We can determine the voltage drop. Then we use V=IR to find each individual current. OR

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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 20Ω 100Ω 40Ω 10Ω ε=1 20V a b i) ii) Point a is at a higher potential. Since current flows from a high potential to a low potential.

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Question Given the following circuit d) Determine the energy dissipated by the 100 Ω resistor in 10 s 10Ω 20Ω 100Ω 40Ω 10Ω ε=1 20V a b Recall: Energy equals Power multiplied by Time

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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 Ωm, determine its radius. 20Ω 100Ω 10Ω 40Ω 10Ω ε=1 20V a b

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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. 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. 10Ω R C r ε S b a

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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. 10Ω R C r ε S b a

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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. 10Ω R C r ε S b a Therefore we need

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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? 10Ω R C r ε S b a Because the top plate is connected to the positive end of the battery, the top is positive. +

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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. 10Ω R C r ε S b a When the current through r is zero, the capacitor is fully charged, with voltage across the plates matching the emf of the battery.

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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. 10Ω R C r ε S b a The current established by the discharging capacitor decreases exponentially.

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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. 10Ω R C r ε S b a

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Question 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. 200Ω ε= 9V 300Ω 400Ω 500Ω

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Question a)Simplify the above circuit so that it consists of one equivalent resistor and the battery. 200Ω ε= 9V 300Ω 400Ω 500Ω R EQ

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Question b) What is the total current through this circuit? 200Ω ε= 9V 300Ω 400Ω 500Ω R EQ =342.2Ω

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Question c)Find the voltage across each resistor. d)Find the current through each resistor. 200Ω ε= 9V 300Ω 400Ω 500Ω VIR 200Ω 300Ω 400Ω 500Ω Total9V0.0263A342.2Ω Let’s use a VIR chart 3.16V A A 5.84V A A

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Question 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. 200Ω ε= 9V 300Ω 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.

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