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a DC Circuits - II (RC) I I a I I R C C R b b RC + + C 2RC C RC - - 2RC q = Ce - t / RC ( q 0 DeSiaMore q =Ce1 -e -t / RC t ) q 0 www.desiamore.com/ifm t 1 Resistor-Capacitor circuits I Let’s try to add a Capacitor to our simple circuit I R Recall voltage “drop” on C? Q V= C Write loop eq: - IR - C Q =0 C What’s wrong here? Consider that I = dQ dt and substitute. Now eqn. has only “Q”: dQ Q -R - =0 dt C Differential Equation ! DeSiaMore www.desiamore.com/ifm 2 Capacitors Circuits, Qualitative Basic principle: Capacitor resists rapid change in Q resists rapid changes in V • Charging (it takes time to change the final charge) – Initially, the capacitor behaves like a wire (V = 0, since Q = 0). – As current starts 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 – Initially, the capacitor behaves like a battery. – After a long time, the capacitor behaves like a wire. DeSiaMore www.desiamore.com/ifm 3 Question The capacitor is initially uncharged, and the two switches are open. E What is the voltage across the capacitor immediately after switch S1 is closed? a) Vc = 0 b) Vc = E Initially: Q = 0 VC = 0 I = E/(2R) c) Vc = 1/2 E Find the voltage across the capacitor after the switch has been closed for a very long time. a) Vc = 0 c)DeSiaMore Vc = 1/2 E b) Vc = E Q=EC www.desiamore.com/ifm I=0 4 Question 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) IR= 0 b) IR=E/(3R) c) IR=E/(2R) d) IR=E/R DeSiaMore E Now, the battery and the resistor 2R are disconnected from the circuit, so we have a different circuit. Since C is fully charged, VC = E. Initially, C acts like a battery, and I = VC/R. www.desiamore.com/ifm 5 RC Circuits (Time-varying currents -- discharging) • Discharge capacitor: C initially charged with Q0 = C Q= Connect switch to b at t = 0. Calculate current and charge as function of time. Q IR + = 0 • Loop theorem C I a b I R + + C - - • Convert to differential equation for Q: Note: Although we know dQ the current is flowing off I =+ the cap., we define it as dt DeSiaMore www.desiamore.com/ifm shown so that … dQ Q R + =0 dt C 6 Discharging Capacitor I a dQ Q R + =0 dt C b I R C • Guess solution: + + - - Q = Q0e - t/ t = C e- t/ RC • Check that it is a solution: Note that this “guess” incorporates the boundary conditions: dQ 1 ₩ - t/ R C ₩ = C e ₩ dt ₩ RC ₩ ₩ dQ Q R dt + = - e t / RC + e t / RC = 0 C DeSiaMore ! www.desiamore.com/ifm t =0 Q =C t= ₩ ₩ Q = 0 7 Discharging Capacitor • Discharge capacitor: I a Q = Q0e - t/ t = C e- t/ RC • Current is found from differentiation: dQ - t/ RC I= =- e dt R Minus sign: Current is opposite to original definition, i.e., charges flow away from capacitor. DeSiaMore b I R + + C - - Conclusion: • Capacitor discharges exponentially with time constant t = RC • Current decays from initial max value (= -/R) with same time www.desiamore.com/ifm 8 constant Discharging Capacitor C Charge on C RC 2RC Q = C e - t/ R C Max = C Q 37% Max at t = RC 0 zero t 0 Current dQ I= = - e- t/ RC dt R I “Max” = -/R 37% Max at t = RC DeSiaMore -/R www.desiamore.com/ifm t 9 Question The two circuits shown below contain identical fully charged capacitors at t=0. Circuit 2 has twice as much resistance as circuit 1. Compare the charge on the two capacitors a short time after t = 0 a) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q2 DeSiaMore Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: t1 = RC and t2 = 2RC. Since t2 > t1 it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor 2 is bigger than www.desiamore.com/ifm 10 that on capacitor 1. RC Circuits (Time-varying currents, charging) I a • Charge capacitor: I R C initially uncharged; connect switch to a at t=0 b Calculate current and C charge as function of time. • Loop theorem Q - IR - = 0 C •Convert to differential equation for Q: I= DeSiaMore dQ dt =R Would it matter where R is placed in the loop?? NO! dQ Q + dt C www.desiamore.com/ifm 11 Charging Capacitor I a • Charge capacitor: I R =R dQ Q + dt C b • Guess solution: C -t Q = C (1 - e RC ) •Check that it is a solution: dQ 1 ₩ - t/ RC ₩ = -Ce ₩ dt ₩ RC ₩ ₩ -t dQ Q - t/ RC R + = e + (1 - e RC ) = dt C DeSiaMore www.desiamore.com/ifm Note that this “guess” incorporates the boundary conditions: t= 0 ₩ Q = 0 t = Q =C 12 Aside… Rather than repeatedly solving the differential equation use the general purpose solution: -t Q = A + Be t or V = A + Be -t t or I = A + Be -t t Where A and B are constants (with the appropriate units) and t is the RC time constant Determine the constants from the boundary conditions at t = 0 and t=infinity. For example, from the last question; t= 0 ₩ Q = 0 t = Q =C DeSiaMore Q (0 ) = 0 = A + B Q ( ₩ ) = C = A www.desiamore.com/ifm 13 Charging Capacitor • Charge capacitor: I a I R Q = C (1- e- t/ RC ) b • Current is found from differentiation: dQ - t/ RC I= = e dt R DeSiaMore C Conclusion: • Capacitor reaches its final charge(Q=C) exponentially with time constant t = RC. • Current decays from max (=/R) with same time constant. www.desiamore.com/ifm 14 Charging Capacitor Charge on C RC 2RC C Q = C (1- e- t/ RC ) Max = C Q 63% Max at t = RC 0 Current dQ - t/ RC I= = e dt R Max = /R t R I 37% Max at t = RC 0 DeSiaMore www.desiamore.com/ifm t 15 Question a • At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. I R b – At time t = t1 =t, the charge Q1 on the capacitor is (1-1/e) of its asymptotic charge Qf = C. – What is the relation between Q1 and Q2 , the charge on the capacitor at time t = t2 = 2t ? (a) Q2 < 2Q1 I C R (c) Q2 > 2Q1 (b) Q2 = 2Q1 • The point is to test your understanding of the time dependence of the charging of the capacitor. • Charge increases according to: Q = C (1 - e -t 2RC ) • So the question is: how does this charge increase differ from a linear increase? • From the graph at the right, it is clear that the charge increase is not as fast as linear. • In fact the rate of increase is just proportional to the current (dQ/dt) which decreases with time. DeSiaMore www.desiamore.com/ifm • Therefore, Q2 < 2Q1. 2Q1 Q2 Q1 Q 16 t t Charging C RC Discharging 2RC C Q = C (1- e- t/ RC ) Q t 0 dQ - t/ RC I= = e dt R I DeSiaMore 0 Q t /R 2RC Q = C e - t/ R C 0 0 RC I -/R dQ - t/ RC I= =- e dt R www.desiamore.com/ifm t 17 t Question I1 This circuit contains a battery, a switch, a capacitor and two resistors I2 I3 C R2 R1 Find the current through R1 after the switch has been closed for a long time. a) I1 = 0 b) I1 = E/R1 c) I1 = E/(R1+ R2) After the switch is closed for a long time ….. The capacitor will be fully charged, and I3 = 0. (The capacitor acts like an open switch). So, I1 = I2, and we have a one-loop circuit with two resistors in series, hence I1 = E/(R1+R2) DeSiaMore www.desiamore.com/ifm 18 Question • At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge. I1 I2 I3 C R2 R1 • What will be the voltage across the capacitor a long time after the switch is closed? (a) VC = 0 (b) VC = R2/(R1+ R2) (c) VC = After a long time the capacitor is completely charged, so no current flows through it. The circuit is then equivalent to a battery with two resistors in series. The voltage across the capacitor equals the voltage across R2 (since C and R2 are in parallel). Either from direct calculation, or remembering the “Voltage Divider Circuit”, VC = VR2 = RDeSiaMore 2/(R1+ R2). www.desiamore.com/ifm 19 Question • At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge. – What is the charging time constant t ? (a) t = R1C (b) t = ( R1 + R2 )C I1 I2 I3 C R1 ₩ R1R2 ₩ (c) t = ₩ C ₩ ₩ R 1 + R2 ₩ • An ideal voltage source contributes no resistance or capacitance time constant is entirely determined by C, R1, and R2. •It might be easier to think about the circuit as if C were discharging; Imagine that the capacitor is charged, and that the battery is replaced by a wire (which also has no resistance or capacitance). Since the battery supplies a constant voltage, it doesn’t affect the time constant. • We need to find the effective resistance Reff through which the capacitor (dis)charges. Looking at the new circuit, it is clear that the capacitor would be (dis)charging through both R1 www.desiamore.com/ifm and R2, which are in parallel their effective DeSiaMore 20 resistance is Reff = R1R2/(R1 + R2) and t = ReffC. R2 Summary • Kirchoff’s Laws apply to time dependent circuits they give differential equations! • Exponential solutions -t t A + Be – from form of differential equation • time constant t = RC – what R, what C ?? You must analyze the problem! SPECIAL CASES • series RC charging solution Q = Q( t = ₩ ) 1 - e- t/ RC ( ) • series RC discharging solution DeSiaMore Q = Q( t = 0) e - t/ RC www.desiamore.com/ifm 21