PHYS 202 Exercise Set #12 W 2010, due March 5 1

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1 PHYS 202 Exercise Set #12 W 2010, due March (6 points) In the circuit shown, the 15-Ω resistor drops a voltage of 4.0 V. (A) Find the battery voltage. You should not have to calculate any currents. (B) Determine the power dissipated by the 40-Ω resistor. (C) How would your answers to (A) and (B) be affected if we removed the 60-Ω resistor? V 60 Ω 40 Ω 30 Ω 15 Ω (A) The parallel 30- Ω and 15- Ω resistor may be combined to an equivalent 10- Ω resistor. We know that the voltage drop across it is 4.0 V. This resistor is in series with the 40- Ω resistor, so they both carry the same current. Since V = I, and I is the same through both, the voltage drop must be proportional to resistance. The 40- Ω resistor has 4 times the resistance of the 10- Ω resistor, so it must drop 4 times as much, or 16 V. The total drop across these two is then 20 V. These are hooked directly to the battery, so the battery voltage must also be 20 V. (B) We use P = V 2 /: (C) They would not be affected at all. P = V 2 = 16 V 40 Ω = 6.4 W

2 PHYS 202 Exercise Set #12 W 2010, due March (5 points) Use Kirchoff s ules to find the voltage V (of the middle battery) needed so that no current flows through the 20-Ω resistor, in the circuit below. I A 16Ω 20Ω V 4Ω B 9.0 V 6.0 V I have labeled the top and bottom nodes A and B in the diagram. If no current flows through the 20- Ω resistor, then there is no voltage drop across that resistor. That means that the battery voltage must be equal to the voltage difference between those two nodes. If we can find the potential of A relative to B, we have the battery voltage V. First we have to find the current I. All the current is around the outside loop. The total voltage is 15.0 V, and the total resistance is 20 Ω. Therefore the current is I = (15.0 V)/(20 Ω) = 0.75 A. Now that we know that, we can find the voltage drop across either of the resistors, which will allow us to get what we want. The drop across the 16- Ω resistor is V = I = (0.75 A)(16 Ω) = 12 volts. I have put in the circuit which side is positive relative to the other. Now, follow the path from node B along the left-hand loop. First we see a 9.0 V rise across the battery, follwed by a -12 V drop across the resistor. This adds up to -3.0 V. Therefore, V = 3.0 V. (The sign is not used when you give a battery voltage. The direction is indicated by the circuit diagram. You may verify for yourself that going around the right-hand loop gives the same answer.

3 PHYS 202 Exercise Set #12 W 2010, due March (5 points) In the circuit below, the voltage drops across bulbs A and C are given. The battery voltage is 14 V. Find the voltage drop across each of the other bulbs. A B C 2 V 1 V 14 V D F E Bulbs A and D are in series and are identical bulbs, so they have the same voltage drop: V D = 2 V. The voltage drops around loop battery-a-d-f must add up to zero. Start at the bottom. There is a 14-V rise, a -2 V drop across A, a -2 V drop across D, and some unknown drop across F. The sum must be zero, so V F = 10 V. (The top side is positive relative to the lower side.) B and C are in parallel, so they must have the same drop: V B = 1.0 V. The drop across the bulbs A and D (together) is 4.0 V. This must be the same as the drop along the path B - E (or along C - E). This is because the A-D combination is in parallel with the BC-E network. Since we know that B drops 1.0 V, E must drop 3.0 V, so that the total is 4.0 V. Summary: D: 2 V B: 1 V E: 3 V F: 10 V

4 PHYS 202 Exercise Set #12 W 2010, due March (5 points) Find the voltages across resistors A, C, E, G, and H in the circuit below. The voltage drops across resistors B, D, F, and J are given. A 12 V B G 4 V E C F D 2 V H J 5 V 4 V We use Kirchoff s loop rule. As an example, look at the loop H-F-J. Start at the node below H, and go clockwise. First a drop or rise V H, then a drop of -2 V across F, then a drop of -4 V across J. These must add up to zero. F J gives a total drop of -6 V, so we must have a rise of 6 V across H. I indicate the signs in the diagram. Now look at the loop C-D-F. Starting in the center node, we have V C, then a -5 V drop across D, then a 2 V rise across F. These must add up to zero, so we must have a 3 V rise across C. We indicate the signs in the diagram. If V C is 5 V, let s look at the B-D-E loop. Starting at the node between B and G, we have a 4 V rise, then a -5 V drop across C, and a rise or drop V E. Clearly V E must be 1 V, with the left side positive relative to the right side. Now we can do the loop G-E-H. Start at the node between B and G and go clockwise. First there is a 1 V rise across E, then a -6 V drop across H, then a rise across G. To make things add up, the rise across G must be 5 V. Now, so find V A, we can use the path A-B-G, A-C-H, or A-D-J, plus the battery. They will all give the same answer. We could have used the path battery-a-d-j right at the first, before we knew the other voltages. Start at the negative terminal of the battery and go clockwise. First a 12 V rise, the a change V A, then a -5 V drop, then a -4 V drop, and we are back. The voltage changes add up to 3 V. Therefore to make the sum zero, there must be a -3 V change (drop) across A. Summary: A: 3 V C: 3 V E: 1 V G: 5 V H: 6 V

5 PHYS 202 Exercise Set #12 W 2010, due March (5 points) In the network shown, some of the resistors have a resistance, and others have resistance. Determine the equivalent resistance between points A and B, in terms of. (Hint: work from the end opposite A and B, using series/parallel reduction. You will quickly see a pattern.) A B Start at the bottom and work up. The bottom parallel pair gives an equivalant resistance. This is in series with the left-hand resistor, giving a sum. This is in parallel with the next resistor, giving resistance again. This combination is in series with the top-left, adding up to. Finally, this is in parallel with the top resistor, giving a total equivalent. A B B A A B ETC.

6 PHYS 202 Exercise Set #12 W 2010, due March (5 points) In the circuit of the previous problem, let = 10Ω. Suppose we hooked an 80.0-volt battery between points A and B. Find the current through the lowest resistor. (That is, the resistor farthest from the battery.) The total current will be (80 V)/(10 Ω) = 8.0 A. This current is called I in the diagram, top left. When it reaches the first node, there are two paths to split into: the resistor and the rest of the network, which also has equivalent resistance. So the current will split evenly. The current continuing below the node encounters the next node, where again there are two paths, both with resistance. It therefore splits in half again. This process continues until we get to the bottom resistor. As the diagram indicates, its current is 1/8 of the original 8 amps, or 1.0 A. V I I/2 I/2 Equivalent resistance I/4 I/8 I/4 I/8 Equivalent resistance Equivalent resistance

AP Physics Electricity and Magnetism #4 Electrical Circuits, Kirchoff s Rules

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