diff options
| -rw-r--r-- | jarrio/week09.bc (renamed from jarrio/week9practice.bc) | 1 | ||||
| -rw-r--r-- | jarrio/week10.bc | 245 | ||||
| -rw-r--r-- | jarrio/week10resistors.py | 66 | ||||
| -rw-r--r-- | li/hw6.tex | 14 |
4 files changed, 320 insertions, 6 deletions
diff --git a/jarrio/week9practice.bc b/jarrio/week09.bc index 611dc8e..b0357f8 100644 --- a/jarrio/week9practice.bc +++ b/jarrio/week09.bc @@ -1,4 +1,3 @@ -######################################################################## print "Weekly Homework 9\n" pi = 4*a(1) epsilon = 8.85 * 10^(-12) diff --git a/jarrio/week10.bc b/jarrio/week10.bc new file mode 100644 index 0000000..80a8fee --- /dev/null +++ b/jarrio/week10.bc @@ -0,0 +1,245 @@ +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) diff --git a/jarrio/week10resistors.py b/jarrio/week10resistors.py new file mode 100644 index 0000000..f7041c7 --- /dev/null +++ b/jarrio/week10resistors.py @@ -0,0 +1,66 @@ +class resistor: + def __init__(self, resistance: float): + self._resistance = resistance; + self.has_current(0) + + def has_current(self, current: float): + self._current = current + self._voltage = current*self._resistance + + def has_voltage(self, voltage: float): + self._voltage = voltage + self._current = voltage/self._resistance + + def __repr__(self): + return f"Resistor with resistance {self._resistance}: current {self._current} through voltage {self._voltage}" + +class parallel(resistor): + def __init__(self, r1: resistor, r2: resistor): + self._r1 = r1 + self._r2 = r2 + self._resistance = 1/(1/r1._resistance + 1/r2._resistance) + + def _update_children(self): + self._r1.has_voltage(self._voltage) + self._r2.has_voltage(self._voltage) + + def has_current(self, current: float): + self._current = current + self._voltage = current*self._resistance + self._update_children() + + def has_voltage(self, voltage: float): + self._voltage = voltage + self._current = voltage/self._resistance + self._update_children() + +class series(resistor): + def __init__(self, r1: resistor, r2: resistor): + self._r1 = r1 + self._r2 = r2 + self._resistance = r1._resistance + r2._resistance + + def _update_children(self): + self._r1.has_current(self._current) + self._r2.has_current(self._current) + + def has_current(self, current: float): + self._current = current + self._voltage = current*self._resistance + self._update_children() + + def has_voltage(self, voltage: float): + self._voltage = voltage + self._current = voltage/self._resistance + self._update_children() + +r24 = resistor(24) +r4 = resistor(4) +r12 = resistor(12) +r5 = resistor(5) +rtot = parallel( r24, series( parallel(r4,r12), r5 ) ) +rtot.has_current(1) +print(r24) +print(r5) +print(r4) +print(r12) @@ -12,8 +12,12 @@ $$P = A(A^TA)^{-1}A^T = \bmatrix{1&1\cr 1&-1\cr -2&4} \fr1{22}\bmatrix{9&4\cr4&3}\bmatrix{1&1&-2\cr1&-1&4} = \fr1{22}\bmatrix{20&6&2\cr 6&4&-6\cr2&-6&20}.$$ -$$Pb = \fr1{22}\bmatrix{20&6&2\cr 6&4&-6\cr2&-6&20}\bmatrix{1\cr2\cr7} -= \fr1{22}\bmatrix{46\cr-28\cr130}$$ +$$p = Pb = \fr1{22}\bmatrix{20&6&2\cr +6&4&-6\cr2&-6&20}\bmatrix{1\cr2\cr7} = +\fr1{22}\bmatrix{46\cr-28\cr130}$$ +With $b = p + q,$ and $p$ known, $q = b-p,$ and $q$ is in the left null +space of $A$ by the definition of orthogonality with the column space +which contains $p.$ \noindent{\bf 7.} @@ -47,12 +51,12 @@ longer distance. \noindent{\bf 16.} -A is a $3\times2$ matrix, $Q$ is also a $3\times2$ matrix, and $R$ is a -$2\times2$ matrix. - $$A = \bmatrix{1&1\cr2&3\cr2&1} = \bmatrix{1/3&0\cr 2/3&1/\sqrt2\cr2/3&-1/\sqrt2}\bmatrix{3&3\cr0&\sqrt2} = QR.$$ +Generally, $A$ is an $m\times n$ matrix, $Q$ is an $m\times r$ (where +$r$ is the rank of $A$) matrix, and $R$ is an $r\times r$ matrix. + \noindent{\bf 17.} $$Pb = QQ^Tb = \bmatrix{1/9&2/9&2/9\cr 2/9&17/18&-1/18\cr |