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11 Cards in this Set

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20.2
A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 19.1 A of current through the torso of a person in 1.83 ms. How much charge moves during this time?
3.49E-2 C
I=deltaq/deltat
I=current (in amps, C/S), q=charge in columbs, t=time

solve for delta Q
SI unit for A, W, Ohm
1 A=1 C/S, 1 W=1 Joule/S=1 kg*m^2/s^3=1 N*m/s, Ohm is (V/A=V/C/S=VS/C)
20.10
Two wires are identical, except that one is aluminum and one is iron. The aluminum wire has a resistance of 0.185 W. What is the resistance of the iron wire?
6.36×10-1 ohm
on page 607 is a table of resistivities. R_aluminum/R_iron=.29. Given resistance_aluminum/.29=6.36E-1.
20.21
The heating element in an iron has a resistance of 26.0 ohm.
The iron is plugged into a 120 V outlet. What is the power delivered to the iron?
554 W.
R and V are given. P=V^2/R. Solve for P. That's it.
20.33
The average power used by a stereo speaker is 40.0 W. Assuming that the speaker can be treated as a 3.90 W resistance, find the peak value of the ac voltage applied to the speaker
17.7 volts
(see page 16 of notebook, page 613 of textbook)
1.Average P=(Vrms)^2/R.
solve for Vrms.
2.Vrms=V_o/sqrrt_2. Solve for V_o. done.
20.48
What resistance must be placed in parallel with a 143 W resistor to make the equivalent resistance 114 W?
5.62×102 ohm
(see page 618, p17 notebook)
1.
1/Rp=1/R_1+1/R_2
Rp=equivalent frequency
equivalent frequency and R_1 are given.
solve for R_2 (hint: you have to divide 1/answer)
20.62
A 56.0 W resistor is connected in parallel with a 123 W resistor. This parallel group is connected in series with a 18.0 W resistor. The total combination is connected across a 12.0 V battery. Calculate the current delivered to the 123 W resistor.
not yet
not yet
Three capacitors (3.82, 5.96, and 12.8 uF) are connected in series across a 55.0 V battery. Calculate the voltage across the 3.82 uF capacitor.
about 25 volts
example is on pdf, electricity and something or other.
here's explanation.
1.you simplify the capacitors' capacitance by doing 1/(sum
of all the reciprocals), should get a number around 2 uF.
That's the C in Q=CV.
2.You plug the answer from [1] into Q=CV, using your
battery's given voltage, and find Q. mine was 108.33 uC.
3.Then you go back to your original capacitor situation
(with three in a series) and solve using V=Q/C, this time
using the capacitance value of the capacitor you're trying
to find the voltage for.

My capacitor's capacitance was 3.82 uF and I ended up with
a voltage of 28.3586.

( also see p 28-29 of problem notebook)
Two cylindrical rods, one copper and the other IRON, are identical in lengths and cross-sectional areas. They are joined, end-to-end, to form one long rod. A 24 V battery is connected across the free ends of the copper-tungsten rod. What is the voltage between the ends of the copper rod?
5.64 V
in a resistor in a series current is the same and is irrelevant, so knock that out. Length and Area are irrelevant as well so all you're left with is the resistivity, and resistivity_1/resistivity_2=voltage_1/voltage_2. figure out the ratio and that's the percentage of 24 that v1 is.
20.62 A 56.0 W resistor is connected in parallel with a 123 W resistor. This parallel group is connected in series with a 18.0 W resistor. The total combination is connected across a 12.0 V battery. Calculate the current delivered to the 123 W resistor.
6.65×10-2 A
see p.38 in problem notebook. Also:
The way I did it, I started with the two parallel resistors
on the left(contains the one whose current you want to
find) and the extraneous resistor on the right.

1.combine the two parallel resistors to get the equivalent
resistance (and collapse the parallel 'box'), you should
now have a box with a battery on one side and two resistors
in a series on the opposite side. (like in example 10 on
page 619)
2.add the resistances together to get the new equivalent
resistance for the entire simplified circuit. (like in
figure 20.24, box a)
3.you should now have a battery and a single resistor
opposite it. Use V=IR and solve for I.
4.Go back to the second box with two resistors in a series
opposite the battery. I is the same across a series, so you
can use the I you got from step 3 and to find the voltage
of the resistor on the left, V=IR again.
5.go back to first circuit. Use the value of [V] you got
from step 4 (second box) and the resistance of the target
resistor given in the problem. V=IR, solve for I. This is
your answer.

If that's not clear just say so.
do number 104 in the text.
for this one say R1=4.07, R2=8.16, R3=5.84, R4=1.98, I1=2.94, V1=12 V
find current for the one with the resistance 1.98.
answer to that series of numbers is 7.04 A, find by drawing everything clockwise, putting the likely drops on left equation likely rises on right, do one for each box (upper and lower_ and one for equation, find currents, do the juncture thing. (just look in the fucking notebok, or at page 627.)