Question
a) In Example 16.8, find the charge on the 8.0F capacitor in Figure 16.21a and the voltage drop across it.
Q = C
?V = V
b) Do the same for the 6.0 F capacitor in Figure 16.21a.
Q = C
?V =
Please also fill in the 3 blanks on the bottom image
a) In Example 16.8, find the charge on the 8.0F capacitor in Figure 16.21a and the voltage drop across it. Q = C ?V = V b) Do the same for the 6.0 F capacitor in Figure 16.21a. Q = C ?V = Goal Solve a complex combination of series and parallel capacitors. Problem (a) Calculate the equivalent capacitance between a and b for the combination of capacitors shown in Figure 16.21a. All capacitances are in microfarads. Figure 16.21 To find the equivalent capacitance of the circuit in (a), use the series and parallel rules described in the text to successively reduce the circuit as indicated in (b), (c) and (d). (b) If a 12 V battery is connected across the system between points a and b, find the charge on the 4.0 muF capacitor in the first diagram and the voltage drop across it. Strategy (a) Using Equations 16.12 and 16.15, we reduce the combination step by step, as indicated in the figure. (b) To find the charge on the 4.0 pF capacitor, start with Figure 16.21c, finding the charge on the 2.0 Mu F capacitor. This same charge is on each of the 4.0 pF capacitors in the second diagram, by fact 5E of the Problem-Solving Strategy. One of these 4.0 pF capacitors in the second diagram is just the original 4.0 pF capacitor in the first diagram.
Explanation / Answer
a)
Q = 2 * 24 = 48 C
V = Q/C = 48/8 = 6 V
b)
Q = C V = 6 * (12-6) = 36 C
V = 12-6 = 6 V
