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course PHY 232
4/29 4:03 pm I know that for a lot these electromagnetism labs that there was always more than we could ever finish in class so I don't have data for every question that's in the QA's.
0110330 Physics II1. Connect the generator in series with a bulb and a long piece of nichrome wire. 'Short out' the nichrome wire by clipping a copper lead to both of its ends, and crank the generator at a constant rate. While cranking, unclip one of the ends of the copper lead and see if you feel a difference in the cranking. Speculate on what the difference tells you and why the circuit behaves as it does.
Report what you observe.
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I don't seem to have written this down but I remember doing it. Shorted out means path of least resistance and more I so more cranking power needed and when going through bulb there is slightly more R and so less I and less cranking power.
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Now remove the copper lead that previously shorted out the circuit. Predict what you will observe if you measure the voltage across the bulb, then across the entire nichrome wire, then across 50 cm of the nichrome wire. After you have committed yourself to your predictions, test them.
Report what you observe.
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Again, no actual data but I predicted above what I think would happen and seem to remember did happen.
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What do you predict you will feel if you allow the nichrome wire to form a loop, crossing itself and making contact at the crossing point?
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This is a short again and will be less R in the short than across the whole loop.
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Replace the bulb with an ammeter, on the highest setting. Again 'short out' the nichrome wire. Be sure you start cranking slowly, and watch the ammeter as you speed up to your original cranking rate (don't exceed the capacity of the ammeter, though). While cranking, remove one of the clips of the 'shorting' lead and see what happens to the ammeter.
Report what you observe.
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Should be less I in the shorted section but of course more I if it's in the other side of the short.
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2. Connect the generator in parallel with the bulb and crank at a constant rate. While cranking attach the two ends of a copper lead, one on each side of the bulb. What happens to the feel of the cranking? Explain what you think is going on to cause this change in the feel.
Report what you observe and give your explanation.
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Less R in the wire than bulb so more I overall than just the bulb.
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Remove the copper lead and replace with the nichrome wire. How does the 'feel' of the cranking change, compared with the change that resulted previous when the copper lead was attached?
Report what you observe.
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I don't recall having done this switch.
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Investigate this circuit with voltmeters and ammeters. Keep careful notes and draw careful diagrams.
Report what you investigated and what you observed.
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Seems I moved straight on to the capacitor here.
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3. Connect the generator in series with the bulb and the capacitor. Crank at a constant rate for about 30 seconds. Describe what happens, how it feels, and what you think is going on.
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Gets easier to turn and bulb slowly dies out. Cap builds resistance to current as I builds in it.
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Sketch the graph you think depicts the current in the circuit vs. clock time.
Describe your graph:
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Concave up decreasing at a decreasing rate approaching 0 as limit.
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Let the capacitor discharge, and repeat, this time with an ammeter in series with the bulb and the capacitor. Observe how current changes with respect to clock time. This means take data so you can sketch a reasonably accurate graph of current vs. clock time.
Describe your graph:
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I have a slightly curve (conave up) line with a few anomalous points between 30-45 seconds. From 120 mA to 25 in 45 seconds.
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Let the capacitor discharge and repeat, this time with a voltmeter in parallel across the capacitor. Observe how the voltage changes with respect to clock time and sketch.
Describe your graph:
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Only data on this is incomplete. In my notes I have 0.2 V @ t=5 and then 0.8, 1.2, 1.6 but with no times and note ""cranked very fast"" Seems that V increased but very quickly I 'died' out and couldn't get very good data for what short time was measurable. I imagine another Ae^-kt type function where k is pretty big.
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Note: I have V vs t data across bulb.
From 1 to a max @ 4 in 50 seconds. Graph was a 'reverse' of I being more curved and concave down. 4. Repeat #3, this time with a resistor in place of the bulb. Answer as on #3:
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This I do have but only for I not V. My data seems to be anomalous as I stayed at 25 mA for 10 seconds, from t = 30 - 40s and then to 20 at t = 45 s. the graph seems to be similar to the I vs t for the bulb but with smaller values of I at first but both seem they would die off similarly given enough time.
@& That does seem to be anomalous but your other data are good, and you know what we expect to happen here.*@
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5. Connect the generator in parallel with the bulb and the capacitor. Investigate this circuit. Take data, sketch diagrams, etc..
Report how you investigated the system, what you observed, what you think it means.
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I drew this circuit and remember setting it up but must have run out of time as I know we did that day so I don't have anything else on this
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University Physics:
6. Using your graph of current vs. clock time in #4, estimate how much charge was transferred to the capacitor.
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It's the are under the curve and using a trapezoidal approximation I get about 1460 milli-Coulombs My later points are a little jumpy so linear is as good as I could do.
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Using in addition your graph of voltage vs. clock time, construct a graph of the rate of change of PE vs. clock time, and use your graph to estimate the PE change of the system:
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Here I looked at my I and V vs t across the bulb since I had both of those and they were fairly consistent. Basically I used corresponding times for both I and V and graphed them with V as x and I as y or I as a function of V and got another Ae^-kt graphs. But I realized after reading the question that it should be IV vs t but I think are under the I vs V graph is Joules/second which is the roc of PE verse clock time. The graph I got for PE vs t is a concave down, not exponential, curve that shows as t goes on PE gets larger then has a max and falls off slightly less quickly than it rose. I hadn't expected it to 'die-off' like it did but the later data points on all of our data seems to be anomalous.
@& Your I vs. V graph is of Coul / sec vs. J / Coul. so your areas would represent Coul / sec * J / coul = J / sec.
You would still need the time information to find the energy.*@
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7. Assume that the capacitor has a voltage proportional to the charge it holds. Based on your data, estimate the proportionality constant.
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Since the graphs are so curved there isn't any one good value but it's clearly related by some inverted value, from reading I know V=Q/C where C is capacitance. I can find this using `d Q= `d I* `d t and then V_ave for the `d t interval but my values aren't anywhere near some 'average' value that would mean much.
@& You might be measuring voltage across the resistor rather than across the capacitor.*@
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8. Assuming this proportionality constant, based on the quantities you have observed, write the differential equation relating dq/dt and q.
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I know R*I= R*dQ/dt= V and now V = Q/c so V_total = R dQ/dt + Q/C Of course it helps for having seen this many times already.
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&#Good responses. See my notes and let me know if you have questions. &#