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print "Daily Homework 10a\n"
print "Question 4\n"
#params
vbattery = 3.05
vchange = 2.9
ammeter = 1.7
bigr = vchange/ammeter
"emf "
vbattery
"internal resistance "
rtot = vbattery/ammeter
rtot - bigr
"circuit resistance "
bigr
print "\nDaily Homework 10b\n"
print "Question 1\n"
#params
r1 = 2
r2 = 5
r3 = 1
r4 = 7
r5 = 3
r6 = 3
r7 = 3
"RA "
r4+r3+1/(1/r1+1/r2)
"RB (open) "
r5 + r4 + r3 + 1/(1/(r6+r1) + 1/r2)
"RB (closed) "
r5 + 1/(1/r4+1/r7) + r3 + 1/(1/(r6+r1) + 1/r2)
print "Question 2\n"
print "Rlong^2\n"
print "Question 3\n"
# params
ratio = 10 # Ip = 10Is
# Without loss of generality, let Is = 1, and R2 = 1, so r = R1.
# IR = V
# 1(1+r) = V
# 10/(1/1 + 1/r) = V
# 1 + r = 10/(1 + 1/r)
# (1+r)(1+1/r) = 10
# r^2 + 2r + 1 = 10r
# r^2 - 8r + 1 = 0
b = 2 - ratio
(-b - sqrt(b^2 - 4*1*1))/2
print "Question 4\n"
#params
ammeter = 9
# clockwise loop rule on top loop
# ammeter - 3*i1 + 2*ammeter = 0
i1 = ammeter
print "I1 (A)\n"
i1
# branch rule
i2 = i1+ammeter
print "I2 (A)\n"
i2
# clockwise loop rule on bottom loop
# -2*ammeter + electromotive - 1*i2 = 0
electromotive = 2*ammeter + i2
print "Electromotive (V)\n"
electromotive
print "\nDaily Homework 10c\n"
print "Question 2\n"
print "C > A = E > B > D\n"
print "Question 3\n"
#params
c = 10 # uF
q = 30 # uC
r = 1.8 # kOhm
qtarget = 15 # uC
# c = q(t)/v(t) --> v(t) = q(t)/c
# dq(t)/dt = -I = -v(t)/r
# dq(t)/dt = -q(t)/cr
# q(t) = Ce^{-t/cr}
# C = q
# q(t) = qe^{-t/cr} = qtarget => -t/cr = ln(qtarget/q)
r = r*1000 # Ohm
qtarget = qtarget/10^6 # C
c = c/10^6 # F
q = q/10^6 # C
t = -c*r*l(qtarget/q)
print "Time (s)\n"
t
print "\nWeekly Homework 10\n"
print "Question 1\n"
#params
r1 = 2.5 # kOhm
r2 = 4.0 # kOhm
r3 = 5.0 # kOhm
v = 100 # V
# v = ir
# p = iv = v^2/r
pparallel = v^2 / (1/(1/r1 + 1/r2 + 1/r3))
pseries = v^2 / (r1 + r2 + r3)
pparallel/pseries
print "Question 2\n"
# I = ef/(r+R)
# P = I^2R = ef^2*R/(r+R)^2
# dP/dR = 0 = d/dR (ef^2*R/(r+R)^2)
# d/dR (R/(r+R)^2) = 0
# (r+R)/(r+R)^3 - 2R/(r+R)^3 = (r-R)/(r+R)^3 = 0
# r = R
print "Part A: R = r\n"
print "Part B\n"
# params
ef = 5 # V
r = 1.8 # Ohm
i = ef/(r+r)
print "Power (W)\n"
i^2*r
print "Question 3\n"
#params
bulbr = 6 # Ohm
r = .8 # Ohm
v = 1.5 # V
iopen = v/(r+bulbr)
print "I_open (A)\n"
iopen
iclosed = v/(r+bulbr/2) / 2 # because current is even between a and b
print "I_closed (A)\n"
iclosed
print "pct change\n"
100*(iclosed-iopen)/iopen
print "Question 4\n"
v = 24 # V
# No params. Effective resistance w/ open is 6Ohm and 6Ohm in parallel,
# giving 3Ohm
print "Part A - Ibat (A)\n"
v/3
print "Part B - dV (V)\n"
v/3
# the two pairs of parallel resistors = two effective resistors in
# parallel
print "Part C - Ibat (A)\n"
v / (1/(1/3 + 1/5) + 1/(1/3 + 1))
print "Part D - dV (V)\n"
0
print "Question 5\n"
v = 24 # V
i6 = v/10 # effective resistance calculation
i5 = v/15 # same
i10 = i5
i4 = i6
print "Currents (i6, i5, i10, i4)\n"
i6
i5
i10
i4
v6 = i6*6 # V = IR
v5 = i5*5
v10 = i10*10
v4 = i4*4
print "Voltages (v6, v5, v10, v4)\n"
v6
v5
v10
v4
print "Question 6\n"
#params
v = 150
rtot = 2 + 1/(1/20 + 1/5) + 4
itot = v/rtot
vparallel = itot / (1/20 + 1/5) # v through the parallel resistors
i20 = vparallel/20
print "I20 (A)\n"
i20
print "Question 7\n"
# 1/(1/4 + 1/12) = 3.
# 3 + 5 = 8
# 1/(1/8 + 1/24) = 6.
# Assume current goes clockwise.
# Loop: 12 - I*6 - 3 - I*3 = 0
# 9 = I*9 => I = 1A.
# Then, week10resistors.py has the solution
print "See source\n"
print "Question 8\n"
#params
tconstant = 7 # ms
print "Charge reduced to half (ms)\n"
-l(1/2)*tconstant
print "Energy reduced to half (ms)\n"
-l(1/sqrt(2))*tconstant
print "Question 9\n"
#params
c = 40 # uF
# The graph shows a reduction from 30V to 10V in 4ms, giving a time
# constant of
tconstant = 4/l(30/10) # ms
# tconstant = cr => r = tconstant/c
c = c/10^6 # F
tconstant = tconstant/1000 # s
r = tconstant/c
print "Resistance\n"
r
print "Question 10\n"
#params
c = .25 # uF
v = 70 # V
r1 = 25 # Ohm (the one the question asks about)
r2 = 150 # Ohm
q = c*v # uC
j = q*v/2
print "Energy dissipated (uJ)\n"
j*r1/(r1+r2)
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