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

2013
Views:

139

Pages:

29
Author:

zorene9q

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 21-3 Energy and Power in Electric Circuit Deriving electric power in a circuit If a charge ∆Q moves across a potential difference V, its electrical potential energy, U, changes by the amount ∆U = (∆Q) V Since the power is the rate of the energy changes with time, we have U (Q)V P  t t Electric Power P=IV (21-4) SI unit: watt, W V2 Also P  , R P  I 2R Problem 21-20 A portable CD player operates with a current of 22 mA at a potential difference of 4.1 V. (a) What is the power usage of the CD players? (b) What is the electric energy the player used in 2 hours of time? Solution: (a) From Eq. (21-4): P  IV   0.022 A  4.1 V   0.090 W (b) The total energy in 2 hours is: U  P  t  0.09  (2  3600)  648 Joules Energy Usage 1 kilowatt · hour (kWh) = (1000 W) (3600 s) = 3.6 x 106 J 1 J= 1/(3.6x106) kWh Problem 21-22 The current in a 120-V reading lamp is 2.3 A. If the cost of electrical energy $0.075 per kilowatt-hour, how much does it cost to operate the light for an hour? Solution: 1. Calculate the power delivered to the lamp: 2. Multiply P by t to find : 3. Multiply by the cost per kilowatthour: P  IV   2.3 A 120 V   280 W U  P t   0.28 kW 1.0 h   0.28 kWh cost   0.28 kWh  $0.075 kWh   $0.021 21-4 Resistors in Series and Parallel Electric circuit often consist of a number of resistors in various ways. To make the circuit simple, a group of resistors can be expressed as an equivalent resistor that has the same resistance for the circuit. Resistors in Series Resistors are connected one after the other, and that they have the same current I through each resistor. Figure 21-6 Resistors in Series Deriving Equivalent Resistance In Fig 21-6 (a), since the total potential difference from point A to point B must be equal to the emf of the battery   V1  V2  V3  IR1  IR2  IR3  I ( R1  R2  R3 ) Similarly, in Fig 21-6 (b), we have   IReq Compare the above two equations, we have Req  R1  R2  R3 Equivalent Resistance for Resistors in Series Req  R1  R2  R3  ...  R SI unit: ohm, Ω Example 21-5 Three Resistors in Series A circuit consists of three resistors connected in series to a 24.0 V battery. The current in the circuit is 0.0320 A. Given that R1=250.0 Ω and R2=150.0 Ω. Find (a) the value of R3, and (b) the potential different across each resistor. Solution Part (a) According to Ohm’s law, the equivalent resistor is  24.0V Req    7.50  102  I 0.0320 A Since Req  R1  R2  R3 we have R3  Req  R1  R2  7.50  102  250.0  150.0  3.50  102  Part (b) Find the potential difference at each resistor, that is current times resistor, respectively V1  IR1  (0.0320 A)(250.0)  8.00 V V2  IR2  (0.0320 A)(150.0)  4.80 V V3  IR3  (0.0320 A)(3.50 102 )  11.2 V Resistors in Parallel Resistors are connected in parallel, and they have the same potential difference. Figure 21-8 Resistors in Parallel Deriving Equivalent Resistance In Fig 21-8 (a), since the total current I is equal to the sum of the current through each resistor, I  I1  I 2  I 3 Since all resistors have the same potential difference,   I1 R1   I 2 R2   I 3 R3 Substitute into the first Eq, we have I  R1   R2   R3  ( 1 1 1   ) R1 R 2 R 3 Now, from Fig 21-8(b), according to Ohm’s law, one has 1 ) I  ( R eq Equivalent Resistance for Resistors in Parallel 1 1 1 1 1     ...   R eq R 1 R 2 R 3 R SI unit: ohm, Ω EXAMPLE 21-6 Three Resistors in Parallel A circuit consists of three resistors, R1=250.0 Ω, R2=150.0 Ω, R3=350.0 Ω and are connected in parallel with a 24.0 V battery. Find (a) the total current supplied by the battery and (b) the current through each resistor. Example 21-6 Three Resistors in Parallel Solution Part (a) Find the equivalent resistor 1 1 1 1    R eq R 1 R 2 R 3 1 1 1     0.01352  1 250.0 150.0 350.0 R e q  (0.01352) 1  73.96  So, the total current is V 24.0V I   0.325 A R eq 73.96 Part (b) Find the currents at each resistor V 24.0V I1    0.0960 A R 1 250.0 V 24.0V I2    0.160 A R 2 150.0 I3  V 24.0V   0.0686 A R 3 350.0 I1  I 2  I 3  I ? Combination Circuits A electric circuit may be more complex, which include resistors both in parallel and series. In this case, we can still applied the equivalent resistors in each part, individually Figure 21-10 Analyzing a Complex Circuit of Resistors Example 21-7 Combination Special In a circuit shown in the diagram, the emf of the battery is 12.0 V, and all the resistors have a resistance of 200.0 Ω. Find the current applied by the battery to this circuit. Example 21-7 Combination Special Solution 1) Find the equivalent resistor of the two resistors Req ,lower  R  R  2 R 2) Find the equivalent resistor of the whole circuit 1 1 1 3    Req R 2 R 2 R Req  2 2 R  (200.0)  133.3  3 3 3) Find the current of the whole circuit I  Req  12.0V  0.0900 A 133.3 Summary Energy and Power in Electric Circuit P U (Q)V  t t P=IV Resistance in Series and Parallel Req  R1  R2  R3  ...  R 1 1 1 1 1     ...   R eq R 1 R 2 R 3 R (21-4) Exercise 21-2 A handheld electric fan operates on a 3.00-V battery. If the power generated by the fan is 2.24 W, what is the current supplied by the battery? Solution Since P=IV, we have IV  P I (3.00V )  2.24W 2.24W I  0.747 A 3.00V Exercise 21-3 Heated Resistance A battery with an emf of 12 V is connected to a 545 Ω resistor. How much energy is dissipated in the resistor in 65 s? Solution 1) Find the power dissipated on the resistor: P = VI = V2 / R = (12V)2/ (546 Ω) = 0.26 W 2) The energy dissipated is ∆U = P ∆t = (0.26 W) (65 s) = 17 J Example 21-4 A holiday goose is cooked in the kitchen oven for 4.00 h. The oven current is 20.0 A and it operates at 220.0 V voltage. The cost of electrical energy is $0.048 per KWH. How much does it cost to cook your goose? Solution 1) Find the power of the oven P = IV = (20.0 A)(220.0 V) = 4.40 x103 W 2) The total energy in 4.00 h is ∆U = P ∆t = (4.40 x103 W)(4.00x60x60 s) = 6.34x107 J =6.34x107 / (3.6x106) kWh =17.6 kWh 3) The total cost is Cost = (17.6 kWh) ($0.048/kWh) = $0.84