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Announcements Quiz II March 3 rd –Median 86; mean 85 Quiz III: March 31st Office Hrs: Today –2-3pm

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+–+– 9 V 5 1.5 V 3 I1 I1 I3 I3 I2 I2 What is the conservation of current law associated with the junction on the right? A)I 1 + I 2 = I 3 B) I 1 + I 3 = I 2 C) I 2 + I 3 = I 1 D)I 1 + I 2 + I 3 = 0 What is the voltage loop rule you get applied to the upper loop? A)9 + 5I 1 + 3I 2 = 0 B)9 + 5I 1 – 3I 2 = 0 C)9 – 5I 1 + 3I 2 = 0 D)9 – 5I 1 – 3I 2 = 0 I 1 + I 3 = I 2 1.5 – 3I 2 = 0 9 – 5I 1 – 3I 2 = 0 I 2 = 1.5/3 = 0.5 A I 1 = (9 – 3I 2 )/5 = 1.5 A I 3 = I 2 – I 1 = 0.5 – 1.5 = – 1 A A Multiloop Circuit

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+–+– 9 V 5 1.5 V 3 I1 I1 I3 I3 I2 I2 9 – 5I 1 -1.5 = 0 I 1 = (9 – 1.5)/5 = 1.5 A A Multiloop Circuit There is one more loop in the problem. We only had one resistor and so only had to consider one current. This can simplify problems!

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+–+– 9 V 5 9 V Odd Circuit What is the current through the resistor? A) 3.6 A B) 1.8A C) 90 A D) 0 A

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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

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RC circuits: Prior to Steady-State +– E C R S1S1 S2S2 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

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RC circuits: Prior to Steady-State +– E R S1S1 C Recall: the voltage across a capacitor is: V=q/C When the capacitor is fully charged the voltage is ( e.g. it acts like a broken wire) Prior, the voltage is V, i.e. there is a voltage drop. Apply the loop rule: Close S 1 The result is a differential equation.

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RC circuits: differential Eqns Differential equation. General Solution: q b and K are determined from boundary conditions and from the parameters of the differential equation Plausibility argument:

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RC circuits: Differential Eqns Integrate both sides to solve: K is determined from boundary conditions Plausibility argument: More general equation and solution:

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RC circuits: Boundary Conditions At t=0, q=0 Charging: As t goes to infinity, q= C Combining these together and: As an exercise do the same for discharging

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RC circuits Capacitor/resistor systems charge or discharge over time Charging: is the time constant, and equals RC. Discharging: 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

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RC circuits: Discharging +– E C R S1S1 S2S2 Circuit with battery, resistor, and capacitor Switch S 1 is closed, then opened At t = 0, switch S 2 is closed What happens? Battery increases voltage on capacitor to V = E At t=0. Current begins to flow Charge Q = C V is stored on capacitor –+ What is the current? [exercise for the class]

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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: .105 .288 .693 2.30 4.60 9.21

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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: 18 V, R = 6, C = 9 µF circuit 3: 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?

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In the figure below, resistor R 3 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 R 3. (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 R 3 goes to infinity. (Think of the wire and bulb quiz ) When R3-> infinity, we can ignore R3

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In Fig. 28N-2a, resistor R 3 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 R 3. (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 R 3 goes to infinity. (Think of the wire and bulb quiz ) When R3-> 0, we can use the loop rule without R2

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