Lecture 11. Kirchoff Circuit Rules RC- Circuits

Transcription

1 Lecture 11 Kirchoff ircuit ules - ircuits

2 1 V 1 1 eview 2 V 2 2 n this circuit, assume V i and i are known. 3 V 3 3 What is 2?? We have the following 4 equations: 1. 2 = V V 3 = V V 2 = V V 1 = 0 Why?? We need 3 NDEPENDENT equations We need 3 equations: which 3 should we use? A) Any 3 will do B) 1, 2, and 4 ) 2, 3, and 4 We must choose Equation 1 and any two of the remaining ( 2, 3, and 4) Equations 2, 3, and 4 are NOT NDEPENDENT Eqn 2 + Eqn 3 = - Eqn 4

3 1 V 1 1 alculation n this circuit, assume V i and i are known. 2 V 2 What is 2?? 3 V 3 2V We have 3 equations and 3 unknowns. 2 = V V 3 = 0 V V 1 = 0 2 V V 2 3 The solution will get very messy in general. nstead do a simpler problem: assume V 2 = V 3 = V; V 1 = 2V; 1 = 3 = with 2 = 2

4 alculation: Simplify n this circuit, assume V and are known. What is 2?? 2V We have 3 equations and V 2 2 = unknowns. V 3-2V V = 0 (outside) -V 2 (2) V= 0 (top) With this simplification, you can verify: 2 = ( 1/5) V/ 1 = ( 3/5) V/ 3 = (-2/5) V/

5 V - V = 0-2V V = = 0 For reference only

6 a 2 b 2V V V Follow-Up licker We know: 2 = ( 1/5) V/ 1 = ( 3/5) V/ 3 = (-2/5) V/ Suppose we short 3 : What happens to V ab (voltage across 2 )? 2V 1 BB (A)V ab remains the same (B)V ab changes sign ()V ab increases (D)V ab goes to zero V cd = +V V bd = +V V ad = V cd = +V Why? edraw: 2 V ab = V ad V bd = V V = 0 a b c V V 2 3 d

7 a b V BB s there a current flowing between a and b? A) Yes B) No At first guess it might appear that since A & B have the same potential urrent flows from battery and splits at A No current flows between A & B Some current flows down Some current flows right

8 licker 1 2 BB (2) = 0 1 = (2) = 0 4 = = 3 1 = 2 3 = 1-3 = = + 3

9 What is the same? urrent flowing in and out of the battery What is different? urrent flowing from a to b

10 What is the same? urrent flowing in and out of the battery What is different? urrent flowing from a to b

11 2 / 3 1 / 3 2 V a b V/2 2 / 3 1 / / 3 12 / 3 1 / / 3 1 / 3 2 / 3

12 licker Problem A B BB c c urrent will flow from left to right in both cases n both cases, V ac = V/2 (2) = 2(4) A = () (2) = () 2(4) B = () (4)

13 ircuits a ε b a ε b ε 2 ε 2 q ( / ) q = ε 1 e t q q = e t / ε 0 t 0 t

14 esistor-capacitor circuits Let s add a apacitor to our simple circuit ecall voltage drop on? Q V = ε Write KVL: ε Q = 0 dq Use = Now eqn. has only Q : dt dq dt KVL gives Differential Equation! ε = 0 We will solve this later. For now, look at qualitative behavior Q

15 apacitors ircuits, Qualitative Basic principle: apacitor resists change in Q resists changes in V harging (it takes time to put the final charge on) nitially, the capacitor behaves like a wire ( V = 0, since Q = 0). As current continues to flow, charge builds up on the capacitor it then becomes more difficult to add more charge the current slows down After a long time, the capacitor behaves like an open switch. Discharging nitially, the capacitor behaves like a battery. After a long time, the capacitor behaves like a wire.

16 licker Problem (two parts) 2A At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. What is the value of the current 0+ just after the switch is thrown? (a) 0+ = 0 (b) 0+ = ε /2 (c) 0+ = 2ε / a ε b 2B What is the value of the current after a very long time? (a) = 0 (b) = ε /2 (c) > 2ε /

17 licker problem (2 parts) 2A At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. What is the value of the current 0+ just after the switch is thrown? (a) 0+ = 0 (b) 0+ = ε /2 (c) 0+ = 2ε / a ε b Just after the switch is thrown, the capacitor still has no charge, therefore the voltage drop across the capacitor = 0! Applying KVL to the loop at t=0+, ε 0 = 0 = ε /2

18 2A licker problem (2 parts) At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. What is the value of the current 0+ just after the switch is thrown? (a) 0+ = 0 (b) 0+ = ε /2 (c) 0+ = 2ε / a ε b 2B What is the value of the current after a very long time? (a) = 0 (b) = ε /2 (c) > 2ε / The key here is to realize that as the current continues to flow, the charge on the capacitor continues to grow. As the charge on the capacitor continues to grow, the voltage across the capacitor will increase. The voltage across the capacitor is limited to ε ; the current goes to 0.

19 licker problem The capacitor is initially uncharged, and the two switches are open. E 3) What is the voltage across the capacitor immediately after switch S 1 is closed? a) V c = 0 b) V c = E c) V c = 1/2 E 4) Find the voltage across the capacitor after the switch has been closed for a very long time. a) V c = 0 b) V c = E c) V c = 1/2 E nitially: Q = 0 V = 0 = E/(2) Q = E = 0

20 licker problem: E 6) After being closed a long time, switch 1 is opened and switch 2 is closed. What is the current through the right resistor immediately after the switch 2 is closed? a) = 0 b) =E/(3) c) =E/(2) d) =E/ Now, the battery and the resistor 2 are disconnected from the circuit, so we have a different circuit. Since is fully charged, V = E. nitially, acts like a battery, and = V /.

21 ircuits (Time-varying currents, charging) harge capacitor: initially uncharged; connect switch to a at t=0 alculate current and a ε b charge as function of time. Q Loop theorem ε = 0 onvert to differential equation for Q: Would it matter where is placed in the loop?? = dq dt dq Q ε = + dt

22 harging apacitor harge capacitor: dq Q ε = + dt Guess solution: a ε b t Q = ε(1 e ) heck that it is a solution: dq dt dq + dt = ε e + Q t / 1 t t / = ε e + ε(1 e ) = ε +! Note that this guess fits the boundary conditions: t = 0 Q=0 t = Q = ε

23 harging apacitor in a circuit harge capacitor: a Q= ε ( t / 1 e ) urrent is found from differentiation: ε b dq ε = = t/ e onclusion: dt apacitor reaches its final charge(q=ε ) exponentially with time constant τ =. urrent decays from max (=ε /) with same time constant.

24 harging apacitor harge on ( t / ) Q = ε 1 e Max = ε 63% Max at t = ε Q 2 dq = = dt Max = ε / urrent ε e t / 37% Max at t = 0 ε / 0 t t

25 Discharging apacitor a dq dt Q + = Guess solution: 0 ε b Q = Q 0e = ε e t / τ t / heck that it is a solution: dq dt = ε / 1 e t e t + = ε / + εe t / = 0! dq dt Q Note that this guess fits the boundary conditions: t = 0 Q = ε t = Q = 0

26 Discharging apacitor Discharge capacitor: a urrent is found from differentiation: Q = Q 0e = ε e dq ε = = e dt t / τ t / t / Minus sign: urrent is opposite to original definition, i.e., charges flow away from capacitor. ε b onclusion: apacitor discharges exponentially with time constant τ = urrent decays from initial max value (= -ε/) with same time constant

27 harge on Discharging apacitor 2 ε Q = ε / e t Max = ε 37% Max at t = Q zero 0 0 t urrent dq = = dt Max = -ε/ ε e t / 37% Max at t = -ε / t

28 licker problem: The two circuits shown below contain identical fully charged capacitors at t=0. ircuit 2 has twice as much resistance as circuit 1. 8) ompare the charge on the two capacitors a short time after t = 0 a) Q 1 > Q 2 b) Q 1 = Q 2 c) Q 1 < Q 2 nitially, the charges on the two capacitors are the same. But the two circuits have different time constants: τ 1 = and τ 2 = 2. Since τ 2 > τ 1 it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor 2 is bigger than that on capacitor 1.

29 licker: The circuit below contains a battery, a switch, a capacitor and two resistors 10) Find the current through 1 after the switch has been closed for a long time. a) 1 = 0 b) 1 = E/ 1 c) 1 = E/( ) After the switch is closed for a long time.. The capacitor will be fully charged, and 3 = 0. (The capacitor acts like an open switch). So, 1 = 2, and we have a one-loop circuit with two resistors in series, hence 1 = E/( )

30 harging Discharging ε 2 ε 2 Q ( t / 1 ) Q = ε e Q Q = ε / e t 0 ε / t 0 0 t dq = = dt ε e t / dq = = e dt ε t / 0 t -ε / t

31 Y&F Problem (Not a clicker problem) A resistor with a resistance of 850 Ω is connected to the plates of a charged capacitor with a capacitance of 4.62 µf. Just before the connection is made, the charge on the capacitor is 8.10 m. A) What is the energy initially stored in the capacitor? Q 1 (8.1x10 ) U = = = 7. 1 J x10 B) What is the electrical power dissipated in the resistor just after the connection is made? V ( Q / ) (8.10x10 / 4.62x10 ) P = = = = 3600 W 850 ) At what time has the energy stored in the capacitor has decreased to half the value calculated in part (A)? U t 1/ 2 = 1 2 t / = Q 2 1/ 2 ln(1/ 1 = U 2 2) = (850)(4.62x = 1 2 )ln(1/ 1 2 Q 2 0 Q 1/ 2 2) = 1.36x10 = 3 s Q 0 2 = Q e 0 t /

32 Power distribution systems onventional House wiring has 240/120V lines Which prong is hot?? (Hint: Black is Ground) emark; houses typically have two single phase, 120V hot lines

33 Electricity osts ($$$$) Your electric costs are charged in units of energy, Kilowatt-hours 1 Kilowatt-Hour = 1000 watts x 3600 sec = 3.6 million joules Oahu residential charges are about 20 cents per K-H. This can be read on your A Kilowatt-Hour meter outside your house. Electricity costs in Hawaii are high! Example 100W light bulb run for 10hrs/day for 30 days uses 0.1 KW x 10 x 30 = 30 Kilowatt-Hours and costs $6.00.