University of California, Berkeley Spring 2012 EE 42/100 Prof. A. Niknejad Problem Set 1 Solutions (Rev B, 2/5/2012) Please note that these are merely suggested solutions. Many of these problems can be approached in different ways. 1. To find the amount of charge, we relate energy to voltage: Q = E V = 500 J 9 V = 55.56 C Since each electron has a charge of 1.6 10 19 C, we need 55.56 1.6 10 19 = 3.47 10 20 electrons. Because we are charging up the battery, we need to move positive charge into the positive terminal. Therefore, electrons are flowing in the opposite direction, from b to a. 2. We can first use dimensional analysis to find the average amount of power consumed over 30 days. $0.16 (30 24 hrs) = $115.2/kW kwh $30 $115.2/kW = 260 W Now we simply apply the power equation to find the average current draw: I = 260 W 120 V = 2.17 A As for your 150 W computer, we already know how much we pay per unit of power consumed, so we can just scale it by 150 W: $115.2/kW 150 W = $17.28 3. We first find the total amount of charge stored on the cell phone battery using dimensional analysis: Q = 1650 mah = (1.65 A) (3600 s) = (1.65 C /s) (3600 s) = 5.94 kc Given this, we can find the energy that the battery is capable of delivering. Since we have the rate at which energy is dissipated (0.4 W), we can also find how long the battery will last. t = E P = Q V P = 5.94 kc 3.7 V 0.4 W = 54945 s = 15.3 hrs
4. The power rating of a light bulb is the power that is produced when connected to the standard line voltage of 120 V. Thus, bulb A with a rating of 35 W has a resistance of R A = V 2 P = (120 V)2 35 W = 411.4 Ω And bulb B with a rating of 85 W has a resistance of R B = V 2 P = (120 V)2 85 W = 169.4 Ω Now consider connecting the bulbs in series. They will have the same current, given by Ohm s law and the series resistance: I A = I B = 120 V (411.4 + 169.4) Ω = 0.207 A Applying Ohm s law to each bulb will give us the voltage: V A = I R A = (0.207 A) (411.4 Ω) = 85 V V B = I R B = (0.207 A) (169.4 Ω) = 35 V Finally, the power across each bulb can be found using any of the above quantities: P A = V A I A = (85 V) (0.207 A) = 17.6 W P B = V B I B = (35 V) (0.207 A) = 7.25 W Since bulb A draws more power, it will be brighter than bulb B. Connecting the bulbs in parallel will leave the same voltage across each one: V A = V B = 120 V Ohm s law can give us the two currents, which are not necessarily equal this time: We can find the power as before: I A = V A R A = 120 V 411.4 Ω = 0.292 A I B = V B R B = 120 V 169.4 Ω = 0.708 A P A = V A I A = (120 V) (0.292 A) = 35 W P B = V B I B = (120 V) (0.708 A) = 85 W Clearly, bulb B will burn brighter in this case. Note that these results should make sense to you; the respective powers agree with the power rating, as we are connecting 120 V across each bulb. 2
5. (a) 0.10 $ /kwhr 1 hr 3600 s 1 kw 1000 J/s = 2.77 10 8 $/J (b) 3.20 $ /gal 264 gal /m 3 1 m3 720 kg 1 kg 4.5 10 7 J = 2.61 10 8 $/J (c) 0.1 A 20 hr 3600 s 1 hr = 7200 C; 7200 C 1.5 V = 10800 J; $0.75 10800 J = 6.94 10 5 $/J Gasoline and residential electric power have almost the same cost per joule; AA batteries are thousands of times more expensive than the other two sources. 6. q = i(t) dt 0 0.085 cos(ωt)e t/τ dt = 0.085τe t/τ (τω sin(ωt) cos(ωt)) τ 2 ω 2 + 1 τ = 0 0.085 τ 2 ω 2 + 1 = 2.69 10 14 C The battery supplies -2.69 10 14 C of charge to the rest of the circuit. The total energy is thus given by E = QV = ( 2.69 10 14 C)(5 V) = 1.35 10 13 J 7. When short circuit happens, large amount of current can flow, which causes excessive heat and a potential fire. The main purpose of a fuse or circuit breaker is to interrupt this flow of current when a short circuit occurs. A fuse disintegrates when it heats up to a certain temperature, therefore opening the circuit in the process. It is a one-time use and would need replacement every time it burns out. A circuit breaker, on the other hand, can be reused multiple times. There are many implementations of a circuit breaker. One implementation is the use of bimetallic strip. When the bimetallic strip heats up, it will bend and displace a certain distance. This bending/displacement can then be used to initiate the switch that trips/opens the circuit. 8. (a) With the two batteries in series with each other and with the load resistor, the circuit looks like the following: 0 3
The current I can be found by writing a KVL loop. Assume we start at the bottom left corner and go clockwise, treating voltage rises as positive and voltage drops as negative: 3 10I + 5 50I 100I = 0 Thus the current through the load is I = 0.05 A. The voltage drop can be found easily using Ohm s law: V = IR = (0.05 A)(100 Ω) = 5 V (b) With the batteries in parallel, the circuit now looks like the following: This time, it is easier to write a nodal equation for the top node and solve for the unknown voltage. Note that this voltage is also the same voltage as v L, the voltage across the load, provided we define the ground on the bottom node as shown. v L 3 + v L 5 + v L 10 50 100 = 0 Thus, the voltage drop across the load is 3.077 V, and the current is (again through Ohm s law) I = V = 0.0377 A. R The current through each battery is the same as the currents through their internal resistances. Thus, the current through the 3 V battery is I = 3.077 3 = 0.0077 A from the positive to negative terminal. On the other 10 hand, the current through the other battery flows in the opposite direction, with a magnitude of I = 5 3.077 = 0.0385 A. 50 (c) A battery rated at 100 ma hr holds a charge of Q = 0.1 C/s hr 3600 s/hr = 360 C Assuming each battery starts off with that amount of charge when hooked up to the load, we can use the current through each battery to find the discharge time, since current is the flow of charge per unit time. Thus, in the series case, both batteries discharge simultaneously: t = Q I = 360 C 0.05 A = 7200 s 4
The parallel case is a bit more complex, since the currents are such that the 5 V battery is charging the other one. Thus, the former must completely discharge before the 3 V battery can start charging. Here we assume that while the first one discharges, the second s charge content remains constant at 360 C (essentially a maximum capacity). t 1 = Q I = 360 C 0.0385 A = 9351 s Once the first battery discharges completely, it becomes a short circuit, and we would need to solve for the new current through the 3 V battery. The new nodal equation is v L 3 + v L 10 50 + v L 100 = 0 This gives us v L = 2.308 V and a current through the battery 0.0692 A. Note that the battery is now discharging, as the current flows in the opposite direction. Thus, the time needed to discharge is t 2 = Q I = 360 C 0.0692 A = 5202 s Note that this is in addition to the time it took to discharge the first battery. Note that the 5V battery is now charging. In a real battery, the terminal voltage would also increase while charging but in our simple model, it would remain at zero until enough charge is transferred to bring it back to full capacity. In this example, during time t 2, a total charge of Q 2 = t 2 I 2 = t 2 v L 50 = 240C is transferred to the second battery. Since this is below the capacity of the battery, in our simple model, it does not recharge to 5V. (d) Connecting two batteries in series is physically viable, and it usually results in an increased total voltage that is just the sum of the two individual voltages (minus any losses from internal resistance). However, connecting them in parallel is often not a good idea unless the cells are well matched. Since the internal resistance is small, this can potentially generate a huge current (consider Ohm s law across the small resistances!), burning out our devices. In the ideal case, the internal resistance tends to 0, giving us a violation of KVL. 9. KCL at the top left node gives us i a = i b = 1 A. At the upper right node, we have i d = i f i e = 1.5 A. Finally, we can calculate i c = i b i d = 2.5 A. Only elements A and B are in series. 10. i e = i a i b = 4 A. i f = i e i d = 7 A. i g = i b + i d i c = 6 A. Here no elements are in series. 5
11. To find voltages, we write KVL equations around the loops in the circuit. For loop DEB, we have 6x x 5 = 0, so x = 1. Thus, v b = 1 V and v e = 6 V. The second loop, ABC, gives us x + v c 5 = 0, so v c = 4 V. No elements are in shunt. 12. Here we immediately see that elements C, D, and F are in shunt, so they must all have the same voltage drop. Thus, v d = v c = 1 V while v f = 1 V. A and E are also in shunt, giving us v e = v a = 3 V. Finally, KVL gives us v a + v b + v c = 0, so v b = 2 V. 6