AP Physics - Problem Drill 14: Electric Circuits

Transcription

1 AP Physics - Problem Drill 14: Electric Circuits No. 1 of Identify the four electric circuit symbols. (A) 1. AC power 2. Battery 3. Light Bulb 4. Resistor (B) 1. Ammeter 2. Resistor 3. AC Power 4. Capacitor (C) 1. Ammeter 2. Battery 3. Light Bulb 4. Resistor (D) 1. AC Power 2. Resistor 3. Light Bulb 4. Battery (E) 1. Ammeter 2. Resistor 3. AC Power 4. Battery A Check the symbol for AC power. Check the symbol for AC power. C. Correct! These are some of the most common symbols that you will see in texts. There may be some slight variation in how people draw them but it usually easy to identify if you know the basics. Check the symbol for resistor. Check the symbol for AC power. These are some of the most common symbols that you will see in texts. There may be some slight variation in how people draw them but it usually easy to identify if you know the basics. The correct answer is (C).

2 No. 2 of A battery has an EMF of 1.5V. However, its terminal voltage is only 1.3 V. Find the internal resistance of the battery when the measured current is 0.30 A. (A) 0.20 Ω (B) 0.67 Ω (C) 0.45 Ω (D) 0.06 Ω (E) None of the above Finding the internal resistance is a bit more complicated than just subtracting the two given voltages. This value does represent the voltage used by the battery itself. B. Correct! The terminal voltage is equal to the EMF minus the lost voltage due to internal resistance. EMF - V lost = V terminal Substitute V = IR for the lost voltage. Solve for the resistance. Use Ohm s law, V=IR, to represent the lost voltage due to internal resistance. The terminal voltage is equal to the EMF minus the lost voltage due to internal resistance. Use Ohm s law, V=IR, to represent the lost voltage due to internal resistance. Known: EMF = 1.5 V Current, I = 0.30 A Terminal Voltage, V terminal = 1.3 V Unknown: Resistance of Battery, R =? Ω Define: The electromotive force, EMF, minus the lost voltage of the battery is equal to the terminal, or remaining, voltage: EMF - V lost = V terminal voltage. According to Ohm s law V=IR. This can be substituted for the lost Output: EMF - IR= T erminal Rearrange to find R Substituting values: 1.5V-.3 A(R)= 1.3 V -.3 A(R)= -.2 V R= -.2 V/-.3 A =.67 Ω This represents the resistance of the battery itself, the internal resistance. Substantiate: Units are correct, sig figs are correct, magnitude seems reasonable. The correct answer is (B).

3 No. 3 of A household outlet of 120 V has a 1000 W hairdryer plugged into it. What is the resistance of the hairdryer? (A).12 Ω (B) 8.3 Ω (C) 14.5 Ω (D) 120,000 Ω (E) 1000 Ω Use P=IV to find the current in the appliance. Then use Ohm s law as a second step. This value is the current, not the resistance. Now use Ohm s law to get the resistance. C. Correct! First use the power formula, P=IV, to find the current in the appliance. Then use Ohm s law, V=IR, to find the resistance. Use P=IV. To find the current, then I=P/V. Don t multiply, divide. Use Ohm s law as V =IR and rearrange to find the resistance. Known: Power, P = 1000 W Voltage, V = 120 V Unknown: Resistance of hairdryer, R =? Ω Define: First, use the power formula to find the current in the hairdryer. Output: P = IV I = P/V Next, use Ohm s law to find the resistance. V = IR R = V/I I = 1000 W/120 V= 8.3 A R = 120 V/8.3 A = 14.5 Ω These two steps could even be combined into one if I=P/V is substituted into ohms law. Substantiate: Units are correct, sig figs are correct, magnitude seems reasonable. The correct answer is (C).

4 No. 4 of The various appliances in your house are wired in what manner? (A) Series, so when one appliance blows a fuse the rest continue to work. (B) Parallel, so when one appliance stops working the rest continue to work. (C) Series, so that the current in all branches of the circuit is constant. (D) Parallel, so that the current in all branches of the circuit is constant. (E) You cannot tell how the house is wired without looking at the circuit. In a series circuit if one appliance blows a fuse, the current cannot flow to the next appliance so all appliances will stop working. B. Correct! Each item in your house gets the same voltage. Additionally, plugging in more items doesn t vary the voltage they get. Even if some items are turned off, or blow a fuse, the rest still function. The current is constant in a series circuit, but this not the type of circuit used. The current is not constant in all branches of a parallel circuit. The voltage is, each item in your house gets the same voltage. If you use your knowledge and observations of how the circuit behaves you can work it out. Do all the appliances get the same voltage? If you add or turn on more and more devices, does this affect the voltage to each of them? If one device is turned off do all the others still work? Consider the properties of the circuits in your house. No matter how many, or which appliances you plug in, they all work on the save voltage. Usually 120V. If one appliance is unplugged, turned off, or blows a fuse, the rest still work. These observation lead to the conclusion that they must be wired in parallel. You may notice that you have a variety of circuits in your house. The kitchen may be on one circuit, the bathroom on another, etc. However, all of these individual circuits are wired in parallel to each other. Therefore, you house is basically a collection of devices wired in parallel. Each of these collections is in turn wired in parallel to other groups. This way, if you overload the kitchen circuit, a fuse or circuit breaker won t disable the entire house, only the section with too much current. The correct answer is (B).

5 No. 5 of Find the electric potential difference of the second resistor, R2, in the circuit shown. (A) 6.75 V (B) 9 V (C).6 V (D) 20 V (E) None of the above 9 V R 1 = 5 Ω R2 = 15 Ω A. Correct! First, simplify the circuit. Next, use Ohm s law, V=IR, to find the current through the circuit. Use Ohm s law once again to find the voltage of the second resistor. First, simplify the circuit. R series = R1 + R2 After the total resistance is found, use Ohm s law to find the current. Once you have the current flowing through this series circuit, use Ohm s law to find the electric potential difference of the second resistor. This could also be called the voltage. First, simplify the circuit. R series = R1 + R2 After the total resistance is found, use Ohm s law to find the current. Known: R 1 = 5 Ω R 2 = 15 Ω V = 9V Unknown: Voltage across V 2 =? V Define: First, simplify the circuit. Since it s a series circuit, simply add the resistors to get the total resistance. R S = R 1 + R 2 Next, use Ohm s law to find the current through the circuit. Since it s a series circuit, this current would be the same throughout. V = IR I = V/R Next, use Ohm s law once again to find the voltage of the second resistor. This value would represent the electric potential difference between the two ends of this resistor. V=IR Output: R s = 5 Ω + 15 Ω = 20 Ω I=9 V/20 Ω I=.45 A V 2 = (.45 A)(15 Ω)= 6.75 V This resistor uses the majority of the voltage from the power supply. This makes sense since it is the largest resistance. Substantiate: Units are correct, sig figs are correct, magnitude seems reasonable. The correct answer is (A).

6 No. 6 of Find the total current flowing through the circuit shown. (A) 3.3 A (B) 33 A (C).33 A (D) 14.9 A (E) 0.67 A 10V R 1 =1Ω R 2 =2Ω First find the total resistance of the whole circuit. Once you have the total resistance of the circuit, use Ohm s law to find the current. You could find the current flowing through each branch of the parallel circuit. Since it s a parallel circuit, the same 10 V is applied to each branch. Sum these currents to get the total current of the circuit. D. Correct! First find the total resistance of the whole circuit. Then use Ohm s law, V=IR, to find the current. The voltage is 10 V from the battery, and the resistance is the total resistance found by combining the resistors in parallel. To get this parallel resistance, the shortcut formula of product divided by sum could be used. First find the total resistance of the whole circuit. Once you have the total resistance of the circuit, use Ohm s law to find the current. Known: R 1 = 1 Ω R 2 = 12 Ω V = 10 V Unknown: Total Current in circuit, I =? A Define: First find the total resistance of the whole circuit = + R R R P 1 2 Next, use Ohm s law, V= IR, to find the total current in the circuit. Output: V=IR I=V/R = + + R P 2 R Ω.67 Ω P 3 I=10 V/.67 Ω I=14.9 A The current in each of the two branches could be found and then combined. This would give the same final answer. Substantiate: Units are correct, sig figs are correct, magnitude seems reasonable. The correct answer is (D).

7 No. 7 of Which of these is an incorrect statement about the circuit shown? R 3 = 50 Ω (A) The three resistors are equivalent to one of 80 Ω. (B) The current in the circuit is 0.25 A. (C) The voltage across R 1 is one 1.25 V. (D) The voltage across R 3 is one tenth of than that in R 1. (E) The voltage across R 2 is five times that in R 1. This is a series circuit so to simplify the circuit, use R S = R 1 + R 2 + R 3. Use Ohm s law to find the current in the circuit, V = IR, where R is the equivalent resistance. To find the voltage across each resistor use Ohm s law, V = IR, the current can be found using the battery voltage and the equivalent resistance. D. Correct! This is a series circuit so we know that the current in each resistor is the same. From Ohm s law V = IR, we see that for a series circuit, if the resistance increases by a factor of 10, the voltage will also increase by a factor of 10. This is a series circuit so we know that the current in each resistor is the same. From Ohm s law V = IR, we see that for a series circuit, if the resistance increases by a factor of 5, the voltage will also increase by a factor of 5. Known: R 1 = 5 Ω R 2 = 25 Ω R 3 = 50 Ω V = 20 V Unknown: Current in circuit, I =? Ω Voltage across R 1, V 1 =? V Voltage across R 2, V 2 =? V Voltage across R 3, V 3 =? V Define: First, simplify the circuit. Since it s a series circuit, simply add the resistors to get the total resistance. R S = R 1 + R 2 + R 3 Next, use Ohm s law to find the current through the circuit. Since it s a series circuit, this current would be the same throughout. V = IR I = V/R Next, use Ohm s law once again to find the voltage of the first resistor. This value would represent the electric potential difference between the two ends of this resistor. V = IR Output: R S = 5 Ω + 25 Ω + 50 Ω = 80 Ω I = 20 V/80 Ω = 0.25 A V 1 = 0.25 A x 5 Ω = 1.25 V V 2 = 0.25 A x 25 Ω = 6.25 V = 5 x V 1 V 3 = 0.25 A x 50 Ω = 12.5 V = 10 x V 1 The correct answer is (D). 20 V R 1 = 5 Ω R2 = 25 Ω

8 No. 8 of 10 needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 8. Students have been asked to find the total current, in the circuit. They all agree the first step is to find the equivalent resistance of the circuit and then use Ohms law to find the total current in the circuit. They cannot agree on the method to use to find the equivalent resistance, who is correct? (A) There is not enough information to find an equivalent resistance. (B) Add all the resistors as if they are in series. (C) Find an equivalent resistance for the parallel resistors, R 3 and R 4, then add this to R 2, finally take that value and use the parallel resistor formula with R 1. (D) Find an equivalent resistance for R 2 and R 4. Find an equivalent resistance R 3 and R 4, finally take that value and use the parallel resistor formula with R 1. (E) Add all the resistors as if they are in parallel. R 2 = 4 Ω R 4 = 6 Ω R 3 = 3 Ω There is enough information to find the equivalent resistance but it will take multiple steps. Check the definition of series and parallel circuits. C. Correct! This circuit contains resistors in parallel and series so you are going to have to use both formulas to find a final equivalent resistor. You are correct that this circuit contains resistors in parallel and series so you are going to have to use both formulas to find a final equivalent resistor, however, R 2 and R 1 are not in series with each other. Check the definition of series and parallel circuits. Step 1: Use = + + R R R eq1 3 4 Step 2: Use R eq2 = R 1 + R eq1 = 4 Ω + 2 Ω = 6 Ω So R eq1 = 2 Ω R eq2 = 6 Ω R 3 = 4 Ω R eq1 = 2 Ω Step 3: Use = + + R R R eq 1 eq2 So R eq = 1.5 Ω R eq3= 1.5 Ω The correct answer is (C).

9 No. 9 of The students have found the equivalent resistance for the circuit, and used Ohm s law to find the total current of 10 A. Now they are asked to find the current I 4, they cannot agree on the next step. (A) The same current flows in all the resistors so the current in I 4 is 10 A. (B) Use Kirchhoff s junction rule to find the current at points 1, 2 and 3. R 4 = 6 Ω (C) Find the equivalent resistance for R 2, R 3, R 2 = 4 Ω and R 4 and use it in Ohm s law to find 3 I 4 the current in the loop 2 and this will be 2 R 3 = 3 Ω the same as current in I 4. (D) Find the equivalent resistance for R 2,R 3, and R 4, and use it in Ohm s law to find the current in the loop 2, use that current to find the voltage across loop 3 containing R 3, R 4, then use Ohms law to find I 4. 1 (E) Half the current goes to loop 1 and half to loop 2, and the current in loop 3 will be half of that in loop 2. So the current is quarter of the total current. Remember this not a series circuit. The current will not be the same in loop 1, loop 2 and loop 3. Kirchhoff s junction rule states that, the current into the junction is equal to the current leaving the junction. It is useful in this case for checking the result but does not tell us what the magnitudes of the currents are. You do need to use the equivalent resistance for R 2, R 3, R 4, to find the current in loop 2, but this is not the same as I 4 since R 3 and R 4 are resistors in parallel, the voltage across them is the same not the current. D. Correct! You do need to use the equivalent resistance for R 2, R 3, and R 4, to find the current in loop 2. This current can be used in Ohms law to find the voltage across the equivalent resistance; this is the voltage across R 3 and R 4 which are in parallel. Ohm s law can then be used to find I 4. Remember Ohm s law, V= IR. Since the resistance in loop 1 is not the same as loop 2 the currents will not be the same. The problem needs to be broken down in to a number of steps. Remember the fundamental equation, Ohm s Law V=IR and at each step carefully consider if the components are in series or parallel. It is important to remember that in a parallel circuit the voltage across each component is the same, this means that the current is not the same in each component. In a series circuit the current that flows through each component is the same so the voltage across each is different. The correct answer is (D).

10 No. 10 of 10 needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 10. Calculate the value for the current I 4 in the resistor R 4. (A) 10 A (B) 2.5 A (C) 1.67 A (D) 5 A (E) 0.83 A 2 R 2 = 4 Ω 3 1 R 4 = 6 Ω I 4 R 3 = 3 Ω This is the total current. Use the information given in questions 8 and 9 to guide you. This is the current in loop 2 but it is not the same as I 4. This is the current in resistor R 3. Use the information given in questions 8 and 9 to guide you. Pay careful attention at each step to whether the components are in parallel or series. E. Correct! Use the information given in questions 8 and 9 to guide you and pay careful attention. Paying careful attention at each step to whether the circuit is in parallel or series. Use the information from questions 8 and 9. Step 1: Find the current in loop 2. This is a parallel circuits so the voltage across R eq2 is the same as that from the power supply,. Using Ohm s law, V = I 2 R eq2 so I 2 = V/R eq2 I 2 = /6 Ω = 2.5 A Step 2: Find the voltage across R eq1 using the current found previously. V 4=I 2 R eq1 = 2.5 A x 2 Ω = 5 V Step 3: Find the current across R 4 using the voltage found previously. R 3 = 4 Ω R eq1 = 2 Ω V 4=I 4 R 4 so I 4 = V 4/R 4 I 4 = 5 V /6 Ω = 0.83 A The correct answer is (E). R 4 = 6 Ω I 4 R 3 = 3 Ω 5 V