University:
California State University, Northridge
Course:
PHYS 100B | General Physics II
Academic year:

2013
Views:

331

Pages:

23
Author:

Rowan Mendoza

Chapter 21 Electric Current and Direct-Current Circuit Outline 21-1 Electric Current 21-2 21-3 Resistance and Ohm’s Law Energy and Power in Electric Circuit 21-4 Resistance in Series and Parallel 21-5 Kirchhoff’s Rules 21-6 Circuits containing Capacitors 21-7 RC Circuits 21-6 Circuits containing Capacitors Capacitors in Parallel Capacitor are connected in parallel, and they have the same voltage. Figure 21-16 shows three capacitors connected in parallel. Figure 21-16 Capacitors in Parallel Deriving equivalent capacitance In fig. 21-16 (a), the magnitudes of the charges on each capacitor are Q1  C1 , Q2  C2 , Q3  C3 , The total charge on the three capacitors is Q  Q1  Q2  Q3   C1   C2   C3   (C1  C2  C3 ) (1) In fig. 21-16 (b), the magnitude of the charge on an equivalent capacitor is Q   Ceq (2) Compared the above two equations (1) and (2), we have Ceq  C1  C2  C3 Equivalent capacitance for Capacitors in Parallel Ceq  C1  C2  C3  ...   C SI unit: farad, F (21  14) Problem 21-56 capacitor in parallel Two capacitors, one 7.5 µF, and the 15µF, are connected in parallel across a 12-V battery. (a) Find the equivalent capacitance of the two capacitors. (b) Find the charge stored in each capacitor. Solution 1) Since the 2 capacitors in parallel: Ceq  C1  C2  Ceq  7.5F  15F  22.5F 2) Both capacitors have the same voltage 12 V: Q1  C  Q1  12(V )  (7.5 10 6 F )  9.0 10 5 C Q2  C  Q2  12(V )  (15 10 6 F )  1.8 10  4 C Capacitors in Series Capacitor are connected in series (one after the other), and they have the same charge. Figure 21-17 shows three capacitors connected in series. Figure 21-17 Capacitors in Series Deriving equivalent capacitance In figure (a), since all the capacitors have the same charge Q, we have V1  Q , C1 V2  Q , C2 V3  Q , C3 And the total potential difference across the three capacitors must equal the battery emf,   V1  V2  V3  Q Q Q 1 1 1    Q(   ) C1 C2 C3 C1 C2 C3 In figure (a), since Q = Ceqε, we have 1  Q (21  16) Ceq Compare (21-15) and (21-16), one has 1 1 1 1    Ceq C1 C2 C3 (21  15) Equivalent capacitance for Capacitors in Series 1 1 1 1 1     ...   Ceq C1 C2 C3 C SI unit: farad, F (21  17) Problem 21-56 capacitor in series Two capacitors, one 7.5 µF, and the 15µF, are connected in series across a 12-V battery. (a) Find the equivalent capacitance of the two capacitors. (b) Find the charge stored in each capacitor. Solution 1) Since the 2 capacitors in series: 1 1 1   Ceq C1 C2  1 1 1    0.20 Ceq 7.5F 15F  Ceq  5.0 ( F ) 1 F 2) Both capacitors have the same voltage 12 V: Q1  Q2  Q  12(V )  (5 10 6 F )  6.0 10 5 C Active Example 21-3 Find Equivalent Capacitance A electric circuit consists of a 12.0 V battery and three capacitors connected partly in series and partly in parallel. Find the total energy store in the capacitors. Active Example 21-3 Find the Equivalent Capacitance and the Stored Energy Solution 1) Find the equivalent capacitance of a 10.0 uf and 5.00 uf in series, 3.33 uF 2) Find the equivalent capacitance of circuit (3.33 uf and 20.0 uf in parallel) Ceq = 23.3 uF 3) Find the stored energy 1 U  CeqV 2  1.68 103 J 2 21-7 RC Circuit A RC circuit consists of a resistor R and capacitor C. Figure 21-18 A Typical RC Circuit Charging the circuit !! The formula that can be used to calculate the charge on the capacitor with time t as a parameter is, q(t )  C (1  e t / ) (21  18), where, e is exponential.  = RC is called Time Constant (s: second), which determines the property of the CR circuit. When t = , q = 0.632 Cε (63.2% of the final charge). Figure 21-19 Charge Versus Time for an RC Circuit The current as a function of time t can be expressed as  I (t )  ( )e t / R (21  19),  = RC is called Time Constant (s: second), Figure 21-20 Current Versus Time in an RC Circuit Example 21-9 Charging a Capacitor A circuit consists of a 126-Ω resistor, a 275-Ω resistor, a 182-uF capacitor, a switch, and a 3.00-V battery all in series. Initially the capacitor is uncharged and the switch is open. At time t = 0, the switch is closed. (a) What charge will the capacitor have a long time after the switch is closed? (b) At what time will the charge on the capacitor be 80.0% of the value found in part (a)? Picture the problem Example 21-9 Charging a Capacitor Solution Time constant   RC  (126  275)(182 106 F )  73.0  103 s Part (a) 1) After a long time ( t  ), q(t )  C (1  et / )  C  (182 106 F )(3.00V )  546 106 C 2) Set q(t) = 0.800 Cε, and solving for t, q(t )  C (1  et / )  0.8C We have 1  et /  0.8 t   ln(0.200) so, t   ln(0.200)  (73.0 103 s) ln(0.200)  118 103 s Summary Equivalent capacitance for Capacitors in Parallel Ceq  C1  C2  C3  ...   C (21  14) Equivalent capacitance for Capacitors in Series 1 1 1 1 1     ...   Ceq C1 C2 C3 C (21  17) RC Circuit q(t )  C (1  et / ) (21  18), Example 21-8 Energy in Parallel Two capacitors, one 12.0 uf, and the other of unknown capacitance C, are connected in parallel across a battery with an emf 9.00 V. The total energy stored in the two capacitors is 0.0115J. What is the value of the capacitance C. Example 21-8 Energy in Parallel Solution 1) Since the Capacitor energy is 1 1 U  QV  CeqV 2 2 2 We have 1 0.0115 J  Ceq (9.00V ) 2 2 Ceq  284 106 F  284  F 2) Solve for C in terms of equivalent capacitance: Ceq  C1  C 284 F  12.0  F  C C  284 F  12.0  F  272  F