1, Describe what power is and write down the equation for electrical power, 2. Write clown the rules for combining resistors and contrast them with the rules for combining capacitors. 3. How can three 10 ohm resistors be put together to give an equivalent resistance of 15 ohms? 4. When measuring voltage drop across a resistor' how do you attach the voltmeter? When measuring current through a resistor, how do you attach the ammeter? 5. Reproduce Ex. 28.10 Analyzing a twoloop Circwt. You don't have to explain qwte so much. 6, Consider Ex, 28,13 Energy dissipated during a capacitor discharge, Describe with words what happens when the switch changes from connecting at a to connecting at b, Aer sometime, also describe what happens when the switch goes back to connecting at a. Compare the time constants of each process. EXAMPLE 28.10 | Analyzing a two-loop circuit Find the current through and the potential difference across the FIGURE 28.31 Applying Kirchhoff's laws. 100 0 resistor in the circuit of FIGURE 28.30. FIGURE 28.30 A two-loop circuit. 12 . Kirchhoff's junction law requires / = 12 + 13- 300 0 100 0 - 200 0 19 V 12 V 300 0 100 4 - 200 ( MODEL Assume ideal batteries and ideal connecting wires. VISUALIZE Figure 28.30 shows the circuit diagram. None of the 19 V 12 V resistors are connected in series or in parallel, so this circuit cannot be reduced to a simpler circuit. SOLVE Kirchhoff's loop law applies to any loop. To analyze a multi- If we substitute ly = 1 - /2 and then rearrange the terms, we find loop problem, we need to write a loop-law equation for each loop. that the two independent loops have given us two simultaneous FIGURE 28.31 redraws the circuit and defines clockwise currents / equations in the two unknowns /, and /2: in the left loop and /2 in the right loop. But what about the middle branch? Let's assign a downward current /, to the middle branch. 400/1 - 100/2 =7 If we apply Kirchhoff's junction law 2/in = Zou to the junction -100/1 + 300/2 = 12 above the 100 0 resistor, as shown in the blow-up of Figure 28.31, We can eliminate /2 by multiplying through the first equation by 3 we see that I, = 12 + /, and thus ly = 1 - 12. If I, ends up being a and then adding the two equations. This gives 1100/, = 33, from positive number, then the current in the middle branch really is which /, = 0.030 A = 30 mA. Using this value in either of the two downward. A negative /, will signify an upward current. loop equations gives /2 = 0.050 A = 50 mA. Because I2 > /, the Kirchhoff's loop law for the left loop, going clockwise from the current through the 100 0 resistor is ]; = 1, -12 = -20 mA, or, lower-left corner, is because of the minus sign, 20 mA upward. The potential difference E(AV); = 19 V-(300 $2)4, - (100 02)/3 - 12 V =0 across the 100 0 resistor is AVicon = /;R = 2.0 V, with the bottom end more positive. We're traveling through the 100 $2 resistor in the direction of /3, the ASSESS The three "legs" of the circuit are in parallel, so they "downhill" direction, so the potential decreases. The 12 V battery is tra- must have the same potential difference across them. The left leg versed positive to negative, so there we have AV = -8 = -12 V. For has AV = 19 V - (0.030 A)(300 02) = 10 V, the middle leg has the right loop, we're going to travel "uphill" through the 100 0 resistor, AV = 12 V - (0.020 A)(100 02) = 10 V, and the right leg has opposite to /3, and gain potential. Thus the loop law for the right loop is AV = (0.050 A) (200 02) = 10 V. Consistency checks such as these are very important. Had we made a numerical error in our E(AV); = 12 V + (100 02)/3 - (200 02)/2 =0 circuit analysis, we would have caught it at this point.CHALLENGE EXAMPLE 28.13 | Energy dissipated during a capacitor discharge The switch in FIGURE 28.39 has been in position a for a long time. There are two ways to determine the energy dissipated in the It is suddenly switched to position b for 1.0 s, then back to a. How resistor. We learned in Section 28.3 that a resistor dissipates energy much energy is dissipated by the 5500 0 resistor? at the rate dE/di = PR = / R. The current decays exponentially as FIGURE 28.39 Circuit of a switched capacitor. 1 = lo exp(-t/T), with lo = AV/R = 9.09 mA. We can find the energy dissipated during a time 7 by integrating: 1200 0 AE = 1Rdi =12R e-25 di = - 471, Re-20 50 V. 200 ALF 5500 0 = TR(1 - e-2Thy The 2 in the exponent appears because we squared the expression MODEL With the switch in position a, the capacitor charges for /. Evaluating for 7 = 1.0 s, we find through the 1200 0 resistor with time constant charge = (1200 02) (2.0 X 10 *F) = 0.24 s. Because the switch has been in position AE = , (1.1 s) (0.00909 A) (5500 02) (1 - e-(2.0s)/(1.Is)) = 0.21 J a for a "long time," which we interpret as being much longer than 0.24 s, we will assume that the capacitor is fully charged to 50 V Alternatively, we can use the known capacitor voltages at / = 0 s when the switch is changed to position b. The capacitor then dis- and / = 1.0 s and Uc = ; C(AVc) to calculate the energy stored in charges through the 5500 0 resistor until the switch is returned to the capacitor at these times: position a. Assume ideal wires. SOLVE Let / = 0s be the time when the switch is moved from a Uc(1 = 0.0 s) = (2.0 x 10-* F) (50 V)2 = 0.25 J to b, initiating the discharge. The battery and 1200 0 resistor are irrelevant during the discharge, so the circuit looks like that of Uc(t = 1.0 s) = } (2.0 x 10 * F) (20 V)2 = 0.04 J Figure 28.34b. The time constant is T = (5500 0) (2.0 X 10 * F) = 1.1 s, so the capacitor voltage decreases from 50 V at / = 0 s to The capacitor has lost AE = 0.21 J of energy, and this energy was AVC = (50 V)e -(10s)(1.Is) = 20 V dissipated by the current through the resistor. ASSESS Not every problem can be solved in two ways, but doing at / = 1.0s. so when it's possible gives us great confidence in our result