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Announcements • Quiz II March 3rd – Median 86; mean 85 • Quiz III: March 31st • Office Hrs: Today – 2-3pm A Multiloop Circuit I1 + I3 = I2 1.5 – 3I2 = 0 9 – 5I1 – 3I2 = 0 I2 = 1.5/3 = 0.5 A I1 = (9 – 3I2)/5 = 1.5 A I3 = I2 – I1 = 0.5 – 1.5 = – 1 A I1 – + 9V 5 I2 3 1.5 V – + I3 What Whatisisthe thevoltage conservation loop rule of you current get applied law associated to the upper with loop? the junction on the right? A) A) 9I1++5II21 =+ I3I 32=0 B) B) 9I+1 +5II13 –=3I I22 = 0 C) C) 9I–2 +5II13+=3I I12 = 0 D) D) 9I1–+5II21 + – 3I I3 2==00 A Multiloop Circuit •There is one more loop in the problem. 9 – 5I1 -1.5 = 0 I1 = (9 – 1.5)/5 = 1.5 A I1 – + 9V 5 I2 3 1.5 V – + I3 •We only had one resistor and so only had to consider one current. This can simplify problems! Odd Circuit What is the current through the resistor? A) 3.6 A B) 1.8A C) 90 A D) 0 A – + 9V 5 9V – + Each of the resistors in the diagram is 12 . The resistance of the entire circuit is: A)120 B) 25 C) 48 D) 5.76 RC circuits: Prior to Steady-State •Thus far we have been referring to circuits in which the current does not vary in time, i.e., steady-state circuits •When we mix capacitors and resistors, the currents can vary with time? Why?! We need to charge the capacitor! •A capacitor which is being charged conducts like a wire •After charging, the capacitor acts like a broken wire E – + S1 C R S2 RC circuits: Prior to Steady-State •Recall: the voltage across a capacitor is: V=q/C E – + S1 •When the capacitor is fully charged the voltage is e ( e.g. it acts like a broken wire) •Prior, the voltage is V, i.e. there is a voltage drop. Close S1 Apply the loop rule: C R q e iR 0 c dq q e R 0 dt c The result is a differential equation. RC circuits: differential Eqns dq q e R 0 dt c q qb Ke t Plausibility argument: dq q R 0 dt c Differential equation. General Solution: qb and K are determined from boundary conditions and from the parameters of the differential equation dq dt q Rc RC circuits: Differential Eqns dq q R 0 dt c dq dt q Rc q Ke t Rc Plausibility argument: Integrate both sides to solve: K is determined from boundary conditions More general equation and solution: dq q e R 0 dt c q qb Ke t RC circuits: Boundary Conditions q qb Ke t Charging: 0 qb K At t=0, q=0 As t goes to infinity, q=eC eC qb Combining these together and: q Ce Cee As an exercise do the same for discharging t RC RC circuits •Capacitor/resistor systems charge or discharge over time Charging: q(t ) Qmax 1 e t / RC Qmax 1 e t / is the time constant, and equals RC. Discharging: q(t ) Qe t / Qualitatively: RC controls how long it takes to charge/discharge completely. This depends on how much current can flow (R) and how much charge needs to be stored (C) As an exercise, show that RC has units sec RC circuits: Discharging •Circuit with battery, resistor, and capacitor •Switch S1 is closed, then opened •At t = 0, switch S2 is closed •What happens? •Battery increases voltage on capacitor to V = E •Charge Q = CV is stored on capacitor •At t=0. Current begins to flow dQ I dt Q(t ) Q0e t /( RC ) What is the current? [exercise for the class] E – + S1 C – + R S2 Time Constants •Time constants are common in science! Given a time constant, t, how long does one have to wait for something to decay by: 10% .105 25% .288 50% .693 90% 2.30 99% 4.60 99.99% 9.21 Four circuits have the form shown in the diagram. The capacitor is initially uncharged and the switch S is open. The values of the emf , resistance R, and the capacitance C for each of the circuits are circuit 1: 18 V, R = 3 , C = 1 µF circuit 2: circuit 3: 18 V, R = 6 , C = 9 µF 12 V, R = 1 , C = 7 µF circuit 4: 10 V, R = 5 , C = 7 µF Which circuit has the largest current right after the switch is closed? Which circuit takes the longest time to charge the capacitor to ½ its final charge? Which circuit takes the least amount of time to charge the capacitor to ½ its final charge? In the figure below, resistor R3 is a variable resistor and the battery is an ideal 18 V battery. Figure 28N-2b gives the current I through the battery as a function of R3. (The vertical axis is marked in increments of 2.5 mA and the horizontal axis is marked in increments of 3.0 .) The curve has an asymptote of 5.0 mA as R3 goes to infinity. (Think of the wire and bulb quiz ) When R3-> infinity, we can ignore R3 V I ( R1 R 2) 18V .005 A( R1 R 2) ( R1 R 2) 3600 In Fig. 28N-2a, resistor R3 is a variable resistor and the battery is an ideal 18 V battery. Figure 28N-2b gives the current i through the battery as a function of R3. (The vertical axis is marked in increments of 2.5 mA and the horizontal axis is marked in increments of 3.0 .) The curve has an asymptote of 5.0 mA as R3 goes to infinity. (Think of the wire and bulb quiz ) When R3-> 0, we can use the loop rule without R2 V I ( R1) 18V .015 A( R1) (R1) 1200 (R 2) 2400