Download presentation

Presentation is loading. Please wait.

Published byLondon Blasdell Modified over 3 years ago

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

3
+–+– 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

4
+–+– 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!

5
+–+– 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

6
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

7
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

8
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.

9
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:

10
RC circuits: Differential Eqns Integrate both sides to solve: K is determined from boundary conditions Plausibility argument: More general equation and solution:

11
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

12
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

13
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]

14
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

15
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?

16
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

17
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

Similar presentations

Presentation is loading. Please wait....

OK

Exam 2: Review Session CH 24–28

Exam 2: Review Session CH 24–28

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google