Physics 9 Fall 2009 Homework 6  Solutions


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1 . Chapter 32  Exercise 8. Physics 9 Fall 29 Homework 6  s How much power is dissipated by each resistor in the figure? First, let s figure out the current in the circuit. Since the two resistors are in series, we can replace them by an equivalent resistance, R eqv = R + R 2 using Kirchhoff s loop law, Vi = E IR IR 2 =. This gives for the current, I = E R +R 2 = 2 = 2 = 2 =.4 amp. now, the power dissipated by each resistor, R i, is P i = I V i = I 2 R i. So, we find for each resistor P = I 2 R = (.4) 2 2 =.92 W P 2 = I 2 R 2 = (.4) 2 8 = 2.9 W.
2 2. Chapter 32  Exercise 27. Determine the value of the potential at points a to d in the figure. Because the wire at point d is connected to a ground, the potential at point d is at V, by definition. In going from point d to point a, e pass through a battery, which bumps up the voltage by V = E = 9 V. So, the potential at point a is 9 V. To get the potentials at points b and c we will need the current in the circuit. From the loop law, V loop = E IR E 2 IR 2 =, where E = 9 V, R = 2Ω, E 2 = 6 V, and R 2 = Ω (note that V across E 2 is E 2 because we re going from positive to negative). Thus, solving for the current, we find I = E E 2 R + R 2 = A. So, the drop across the first resistor is V = IR = 2 V, which means that the potential at point b is 92 = 7 V. Next, the battery connected backwards fights the 7 V, dropping it to 76 = V. So, the potential at point c is V. Finally, we cross the last resistor which drops the potential by V = IR 2 = V, giving an net potential of  = V at point d, which we know. So, altogether, Point a b c d Voltage 9 V 7 V V V 2
3 3. Chapter 32  Exercise 34. What value resistor will discharge a. µf capacitor to % of its initial charge in 2. ms? For a capacitor of initial charge Q connected to a resistor, R, the charge falls away exponentially fast, Q (t) = Q e t/rc. So, a capacitor will fall to % of it s initial charge after a time t such that Q (t ) =.Q = Q e t /RC. = e t /RC. Taking the natural logs of both sides and rearranging (using the fact that ln. = ln ) gives Plugging in the numbers gives R = t C ln. R = t C ln = ln =.87kΩ. 3
4 4. Chapter 32  Problem 37. The figure shows six identical bulbs connected to an ideal battery. All the bulbs are glowing. Rank in order, from brightest to dimmest, the brightness of bulbs A to F. Explain. The brightness of the bulbs depends on the power dissipated in the bulb, which depends on the current. So, whichever bulb has the highest current has the highest power dissipated, and so is the brightest. Now, because bulb A is in parallel with the rest of the bulbs, the current from the battery is split between bulb A, and all the rest of the bulbs! So, A is brightest. The current next passes through bulb D before splitting up to the rest of the bulbs, and so D is next brightest. Next, the current passes through the next set of bulbs. Bulbs C and E are in parallel with each other, so they are equally bright. The current through F is the sum of the currents through C, E, and B, and so F is brighter than C and E. Now, B is in parallel with C, E, and F. But, since the CEF combination has a higher resistance than B, more current passes through B than the others since they are all at the same potential. So, we have for brightness, A > D > B > F > C = E. 4
5 5. Chapter 32  Problem 4. You have three 2 Ω resistors. Draw diagrams showing how you could arrange all three so that their equivalent resistance is (a) 4. Ω, (b) 8. Ω, (c) 8 Ω, and (c) 36 Ω. The four diagrams are seen in the figure below. (a) In the first diagram, all three resistors are in parallel. So, the equivalent resistance is R eqv = R + R 2 + R 3 = = 4, which gives R eqv = 4 Ω. (b) In this case the top two resistors are in series and are both in parallel with the bottom resistor. The top resistors add to give R + R 2 = = 24. This equivalent resistance combines with the resistor on the bottom to give and so R eqv = 8 Ω. R eqv = = 3 24 = 8, (c) Next, we have a two resistors in parallel in series with a third resistor. The two in parallel combine to give R eqv = +, giving an equivalent resistance of 6 Ω for 2 2 that branch of the circuit. Adding this in series with the other resistance gives R eqv = = 8 Ω. (d) Finally, we just have three resistors in series, which all add together to give R eqv = R + R 2 + R 3 = = 36 Ω. 5
6 6. Chapter 32  Problem 42. What is the equivalent resistance between points a and b in the figure? We can form an equivalent circuit in the following steps. In the first step, we break up the three parallel resistors into a single equivalent resistor. Since they are in parallel, R eq = + + = 6, and so R eq = 2 Ω. Next, the element that contains the Ω resistor in parallel with the wire has a net resistance of zero! Why? The current will take the path of least resistance, bypassing the resistor altogether. (The current doesn t take a single path, in general  it splits between the two paths depending on the load resistance, as we ll see). This circuit is a short circuit. Can we see this? Think of the wire as a resistor with R =. Then, for the resistor and wire in parallel, R eq = R + R 2 = + =. So, R eq =. So, we can just ignore the Ω resistor to get the three resistors in series in the third diagram. This gives a net resistance of R eq = = 7 Ω. 6
7 7. Chapter 32  Problem 49. (a) Load resistor R is attached to a battery of emf E and internal resistance r. For what value of the resistance R, in terms of E and r, will the power dissipated by the load resistor be a maximum? (b) What is the maximum power that the load can dissipate if the battery has E = 9. V and r =. Ω? (c) Why should the power dissipated by the load have a maximum value? Explain. (a) When the battery is attached to the load resistor, R, the internal resistance, r, of the battery is in series with the external resistor. From our usual loop law, the current in the wire is I =. The power dissipated in the load resistor is P = I 2 R = the derivative gives E R+r E2 R. The power dissipated is a maximum when dp (R+r) 2 dr ( ) ( ) d E 2 R dr (R + r) 2 = E 2 (R + r) 2 2R (R + r) (R + r) 4 = =. Taking Solving gives (R + r) 2 2R (R + r) = R 2 r 2 =, and so the power dissipated is a maximum when R = r. (b) The power is maximized when R = r, which gives P = E2 R (R+r) 2 = E2. So, plugging 4r in the numbers gives P = 9.2 = 2.25 W. 4 (c) Suppose R is very small. Then the current is big, I E/r. But, P = I 2 R, which is forced to be small by the factor of R. So, the smaller R gets, the smaller the power goes. What if R is big? In this case, the current is also small, I E/R. Then, the power becomes P = I 2 R E 2 /R, which is driven to zero as R gets big. So, in either case the power is driven down for either extreme, R, or R. There has to be some intermediate value of R that maximizes the power dissipated, and this is what we found in part a. 7
8 8. Chapter 32  Problem 63. For the circuit shown in the figure, find the current through and the potential difference across each resistor. Place your results in a table for ease of reading. First, we need to determine the current being supplied by the battery. We begin by simplifying the circuit as in the following diagram, applying usual rules for combining resistors in parallel and in series. To move from the first to the second diagram, we combine the 4 Ω and 2 Ω resistors in parallel for a net resistance of 3 Ω. Then we combine the 3 and 5 Ω resistors in series to get the 8 Ω resistors in the third diagram. Next, combine the 24 and 8 Ω resistors in parallel to get the 6 Ω resistor in the fourth diagram. Finally, we combine the 6 and 3 Ω resistors in series to get the equivalent 9 Ω resistor in the last circuit Applying Kirchhoff s loop law to the equivalent circuit gives E IR E 2 =, where the voltage drops moving across the second battery since the current is moving from positive to negative. This gives a current, I = E E 2 = 2 3 = amp. R 9 So, now we need to work backwards, reconstructing the circuit as seen in the following diagram. 8
9 We begin with step. Because current is conserved in the last diagram, we know that A flows through both the 3 and 6 Ω resistors. The drop in voltage across the 3 Ω resistor is thus V = IR = 3. V. The change in voltage across the equivalent 6 Ω resistor is V = IR = 6 V. Step 2  We split up the equivalent resistor into the two original resistors. We know that the change in voltage across each resistor is the same, and is equal to 6 V, but the current, which splits between the two resistors, is not the same in each resistor. From Ohm s law, we know that V = IR, and so I 24 = 6/24 =.25 A, while I 8 = 6/8 =.75 A. Notice that I 24 + I 8 = A, which we need. Step 3  Now we break up the equivalent 8 Ω resistor into the two resistors in series. Now the current is the same through each of these two resistors, but the voltage is not. Again, going back to Ohm s law, we can solve for the change in voltage through each resistor. V 3 =.75 3 = 2.25 V, while V 5 =.75 5 = 3.75 V. Again, notice that the voltages in these two resistors adds up to 6 V, which we knew was the case. Step 4  Finally, we break up the 3 Ω equivalent resistor into the 4 and 2 Ω resistors in parallel. Now, the current through these two resistors aren t equal, but the voltage drop is. So, yet again, we go back to Ohm s law. I 4 = 2.25/4 =.56 A, while I 2 = 2.25/2 =.9 A. Again, the currents add up to the original.75 A, as it must. So, now we have all our answers. Let s put the all in a table Resistor (Ω) Voltage (V) Current (A)
10 9. Chapter 32  Problem 77. The capacitor in Figure 32.38a begins to charge after the switch closes at t = s. Analyze this circuit and show that Q = Q max ( e t/τ ), where Q max = CE. When the switch closes, the battery begins to charge. Once it starts, Kirchhoff s loop law says that E IR q C = I = E R q RC. But, I = dq dq (the derivative is positive since the capacitor is charging). So, = E q dt dt R dq Separating and integrating gives = dt, or q CE RC Q Solving for Q (t) gives dq q CE = RC t ( ) Q CE dt ln CE Q (t) = EC ( e t/τ), = t RC. where τ RC. So, we can write Q = Q max ( e t/τ ), where Q max EC. We know that this is Q max because it is the maximum charge that builds up after a long time (t ). RC.
11 . Chapter 32  Problem 78. The switch in Figure 32.38a closes at t = s and, after a very long time, the capacitor is fully charged. Find expressions for (a) the total energy supplied by the battery as the capacitor is being charged, (b) total energy dissipated by the resistor as the capacitor is being charged, and (c) the energy stored in the capacitor when it is fully charged. Your expressions will be in terms of E, R, and C. (d) Do your results for parts a to c show that energy is conserved? Explain. In problem 77 we found that Q (t) = CE ( e t/rc) for this circuit. We ll use this in what follows. (a) The power being supplied by the battery as the capacitor is being charged is given by P batt = I V batt = IE. Now, as the capacitor is being charged, I = dq dt, or I = d dt [ CE ( e t/rc )] = E R e t/rc. So, P batt = IE = E2 R e t/rc. Now, the power is how fast the energy is changing, P batt = de. So, we can integrate the power to get the total energy supplied by dt the battery. E batt = P batt dt = E 2 dt e t/rc = E 2 C C RCe t/rc = CE 2, where we integrated to infinity because the capacitor has been charging for a long time. (b) Now, the power dissipated by the resistor is P res = I V res = I 2 R, using Ohm s law. So, plugging in the expression for the current gives P res = E2 R e 2t/RC. But again, P res = d E dt res. So we again integrate to find the total energy. E res = P res dt = E 2 dt e 2t/RC = E 2 C R 2 e 2t/RC = 2 CE 2. So, the total energy dissipated by a resistor is E res = 2 CE 2. (c) The total energy stored by a capacitor when it has charged to Q max = CE is E cap = Q2 max 2C = (CE)2 2 = 2 CE 2. (d) The total energy supplied by the battery is E = CE, and is the only source of energy. The energy that the capacitor stores up is 2 CE 2, while the resistor dissipates a total energy of 2 CE 2. So, the total energy stored by the capacitor, plus the energy dissipated by the resistor is 2 CE CE 2 = CE 2, which is precisely the energy supplied by the battery. So, the total energy is conserved!
= (0.400 A) (4.80 V) = 1.92 W = (0.400 A) (7.20 V) = 2.88 W
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