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BIOE 3010 University of Toronto Kirchhoff Laws Questions

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Exam 1 BIOE3010
Please show all work for partial credit. Each question is worth 10 points.
1. Use Kirchhoff’s Laws to find the current through each resistor. Show all work.
2. For the circuit below – Name it, then given Vin = -5V, R1 = 330Ω, R2 = 5kΩ, find Vout and the current
through the circuit – draw in the current direction.
3. Reduce resistors
a. Find the equivalent resistance and current through each of the resistors.
b. Find the equivalent resistance
To defibrillate a heart safely, the right amount of current should be used. If the chest has a resistivity of
170Ω and the potential difference applied is 1040 Volts, what is the current flowing through the heart for
a few milliseconds?
BIOE 3030: Introduction to
2. Chemical Structure of Biomaterials
2.2. Structure of Metals
2.2.1. Crystal Structure

Coordination number: the number of nearest-neighbor atoms.

Atomic packing factor (APF): a means to estimate how much
occupied space there is in a given volume.
APF = volume of atoms in a unit cell / total unit cell volume
*unit cell: the configuration of atoms in a small section of crystal that is
repeated. Face-centered cubic (FCC) structure
• Atoms are located at each of the corners and the
centers of each face.
• Coordination number: 12
• APF: ? Body-centered cubic (BCC) structure
• Atoms are located at all eight corners and a single
atom at the center.
• Coordination number: 8
• APF: ?
Hexagonal close-packed (HCP) structure
• Atoms are located at all twelve corners, three atoms
at the center, and two atoms at top and bottom.
• Coordination number: 12
• APF: ?
2.2.2. Crystal system
Generic lattice structure
Edge length: a, b, c
Angles: α, β, ϒ
Miller indices
• Indicate the location of points and the orientation of
planes in the lattice structure.
Coordinates of atoms
• are written as h, k, l
• three indices represent fractions of the lattice
parameters, a, b, and c.
Miller indices for planes
Determine the points at which the plane intersects x, y, and z axes. If the
plane is parallel to an axis, the intercept for that axis is taken to be .
Take the reciprocal of the intercepts.
Multiply by an integer to clear fractions.
Record integer indices in parentheses with no commas (h k l).
Negative numbers are indicated by a bar over the integer.
What is,
1. free radical polymerization?
2. semi-crystalline structure?
3. stress-strain curve?
4. degradation mechanism of metal and polymer?
5. contact angle analysis?
6. protein adsorption mechanism?
7. mRNA and tRNA?
8. inflammatory response mechanism?
BIOE 3030: Introduction to
2.2.3. Defects in Crystal Structures Point defects
• A vacancy is found at a lattice site where an atom would
normally be present, but is currently missing.
• A self-interstitial occurs when an atom from the crystal is
crowded into the interstitial space between two adjacent
2.2.4. Solid state diffusion
• The movement of material via atomic motion.
• Since at room temp most metals are solids, the type of diffusion that
occurs through metals is solid state diffusion.
• Vacancy diffusion (a): an atom jumps to an adjacent vacancy, thereby
exchanging the location of the atom and the vacancy.
• Interstitial diffusion (b): an atom migrates from one interstitial
position to a neighboring position. Modeling of diffusion
2.3. Ceramics
2.3.1. Crystal Structure

is affected by two parameters: the magnitude of the electrical charge of
ions and the physical size of these ions.

The electrical charge is important because the crystal must remain
electrically neutral.

The physical size of ions is the knowledge of the radii of both the cations
(rc) and anions (ra).

Coordination number: the number of nearest neighbors with opposite
charge. AX Crystal Structures

“A” represents the cation.

“X” represents the anion.
coordination number: 6
coordination number: 8 AmXp crystal structures

Ceramics are often composed of cations and anions that do not have equal

“m” and/or “p”  1.
2.4. Structure of Poymers
2.4.1. General structure Repeat Units Molecular Weight Determination
Number average molecular weight (Mn)
Weight average molecular weight (Mw)
2.3. Mw/Mn
• Polydispersity index, PI
• It is used as an indicator of the breadth of the molecular
weight range in a polymer sample.
• If the ratio is close to 1, the sample has monodisperse
molecular weight range.
• If the ratio has larger numbers, the sample has polydisperse
molecular weight range.
Example 1:
Suppose we have a polymer sample consisting of 9 moles having
molecular weight 30,000 and 5 moles having molecular weight
50,000. Calculate PI.
Example 2:
Suppose our sample consists of 9 grams having molecular weight
30,000 and 5 grams having molecular weight 50,000. Calculate PI.
Suppose our sample consists of A: 5 grams having molecular
weight 100 and B: x grams having molecular weight 50, and the
PI of this polymer sample is 1.07. Obtain the quadratic equation
to calculate the weight (x) of polymer B.
BIOE 3030: Introduction to
Biomaterials AmXp crystal structures

Ceramics are often composed of cations and anions that do not have equal

“m” and/or “p”  1.
2.4.2. Polymer Synthesis
Vinyl polymerization
• Free radical polymerization
• Polymerization occurs through reactive end of growing chain.
• A radical initiates the polymerization.
R● + CH2=CH2 –>
RCH2CH2● + CH2=CH2 –>
+ nCH2=CH2 –>
R(CH2CH2)n+1CH2CH2● + R● –>
R(CH2CH2)n+1CH2CH2● + R(CH2CH2)m+1CH2CH2● –>
Peroxide (ROOR) and hydroperoxide (ROOH)
• Thermally unstable.
• ROOR  two alkoxy radicals (2RO•).
• ROOH  alkoxy radical (RO•) + hydroxy radical (•OH).
Azo compounds
• Compounds having cyano group (-CN).
• Thermally unstable.
• It produces cyanoalkyl radical and nitrogen gas.
• Unstable under UV light.
• Polymerization at low temperature.
You have following chemicals. You want to synthesize a polymer
by free radical polymerization at room temperature. Choose
proper two chemicals (initiator and monomer) and synthesize
the polymer. Draw complete reaction in each step of initiation,
propagation, and termination.
Non-vinyl polymerization
Synthesis of polyesters
Synthesis of polyesters by direct esterification
• A reaction between a carboxylic acid, RCOOH, and an alcohol,
R’OH, forming an ester, RCOOR’, and water.
Synthesis of polyesters by trans-esterification
• A reaction (or exchange of alkyl chains) between an ester, RCOOR’,
and an alcohol, R”OH, forming another ester, RCOOR”, and
another alcohol, R’OH. Synthesis of polyesters by reaction between alcohol and acyl
• Acyl chloride, RCOCl, reacts with an alcohol, R’OH, and forms an
ester, RCOOR’, and hydrochloric acid, HCl
BIOE 3030: Introduction to
2.5. Techniques: Introduction to Material Characterization
2.5.1. X-Ray diffraction
• Examines how X-rays are diffracted from atoms in a material. Basic Principle
• Diffraction of X-rays occurs when incident rays are scattered by atoms in a way that
reinforces the waves.
• If two waves (wave 1 and 2) of the same wave length (λ) and amplitude (A) are
in phase at point O/O’ and are scattered so that they remain in phase (wave 1’
and 2’), they will combine in a reinforcing manner and the diffracted wave will
have the same wavelength, but twice the amplitude (2A).
• This occurs only if the total distance traveled by waves 1 and 2 differ by an
integer number of wavelengths.
• When the path lengths of two incident waves (wave 3 and 4) differ by a number of
half wavelengths, after scattering, the waves (wave 3’ and 4’) are out of phase and
there is destructive interference.
• The scattered waves cancel each other and no diffraction occurs.
• When X-rays enter a sample, they are scattered in all directions.
• In most cases, this leads to destructive interference and no resulting X-rays can be
• For certain atomic arrangements, diffraction occurs and X-ray patterns can be
The planes of atoms, A/A’ and B/B’ have the same Miller indices (h k l) and are separated by
the distance dhkl. Waves 1 and 2 are parallel X-rays of wavelength λ that are in phase upon
striking the sample at angle φ. Wave 1 is scattered by atom P, and wave 2 is scattered by
atom Q. For diffraction to occur, the total path length of the two waves must be integer
number of wavelengths.
2.5.2. Ultraviolet and visible light spectroscopy (UV-VIS) Basic Principle
The absorption of UV-VIS radiation by a molecule (M) may promote one or more of its
valence electrons to a higher energy state, resulting in molecular excitation to a new state
ν: frequency of radiation
h: Planck’s constant (6.6 x 10-34 J/s)
• Usually, it involves the transition
of electrons from σ or π bonding
to antibonding molecular orbit.
• Energy absorption occurs at
various wavelengths, depending
on the chemical structure of the
2.5.3. Infrared Spectroscopy (IR) Basic Principle
• In order for interaction to occur, the bond must possess a permanent dipole.
• A polar bond can be thought of as oscillating or vibrating with a certain frequency.
• If the frequency of the IR radiation matches the frequency of this vibration, then there is
a reinforcing interaction and the amplitude of the oscillation increases.
2.5.4. Nuclear Magnetic Resonance Spectroscopy (NMR)
• Like little magnet, when the nuclei are placed in a larger, external magnetic field, their
orientation can be in the same direction as the field, which is more energetically
favorable and thus represents a lower energy state, or opposite to the field, which
represents a higher energy state.
α: lower energy state β: higher
energy state
• The difference in energy ΔE
between these two states is
equal to hv.
• If exposed to radio energy of
frequency v (resonance
frequency), then the nuclei
will absorb this energy and
the nuclear spin will be
flipped from the α to β
state. At this point, the
nuclei are in resonance.
• In addition to the external magnetic field, the
1H nucleus is also affected by smaller magnetic
field created by the circulation of electrons
around it. The movement of these electrons
creates a magnetic field whose strength is
proportional to the electron density. This
smaller field opposes the effects of the
external field.
• In the case of the H-C bond, there are more
electrons nearby than would normally be
found in a hydrogen atom, so the nucleus is
said to be shielded from the strength of the
external field. Therefore, more energy is
needed to induce resonance in this molecules.
• On the other hand, if the H is bonded to a
more electron-withdrawing atom such as F, the
electron density around the H nucleus is less
than normal, and the nucleus is deshielded. In
this case, less energy is required than normal is
BIOE 3030: Introduction to
4. Mechanical Properties of Biomaterials
4.2. Mechanical Testing Methods, Results and Calculation
Tensile Testing
From American Society for Testing and Materials (ASTM) standard, the tensile sample is
shaped into a “dog-bone” geometry.
In tensile testing, engineering stress (σ)
F: Force applied perpendicular to the cross-section of the sample
A0: original cross-sectional area of the sample
The unit of stress: Pa
Engineering strain (ε)
l0: length of the sample before loading
li: length of the sample at any point during the testing procedure
Compression Testing
• The same equation are used to calculate stress and strain as for
tensile tests, but since the compressive force acts in the opposite
direction to the tensile force, F is negative, resulting in a negative
• Additionally, since the sample becomes smaller along the axis of
the stress, l0 is larger than li, and the calculated strain is negative.
Shear Testing
• Shear testing produces forces that are parallel to the top and
bottom faces of the samples. Shear stress (τ) can be calculated by
F: force imparted parallel to the upper and lower faces
A0: area of these faces
• Since the shear force causes sample deformation of angle θ,
shear strain (ϒ) is defined as
(4.4) Stress-Strain Curves and Elastic Deformation
• Looking at Material I, it can be seen that the stress and strain are proportional to each
other at all values. This relationship is known as Hooke’s law.
E: modulus of elasticity or Young’s modulus (MPa)
• In the area of the curve where the stress and
strain follow this linear relationship, the
sample undergoes elastic deformation.
• Elastic deformation is not permanent; upon
release of the load, the sample returns to its
original shape.
• Shear stress and strain can be related by
• G is the shear modulus, which represents
the slope of the stress-strain curve in the
elastic region. Stress-Strain Curve and Plastic
• In contrast to elastic deformation, plastic
deformation is permanent. For example,
Materials II,III, and IV in the nonlinear
portion after the elastic region.
• The beginning of plastic deformation is
the point where the stress-strain
relationship no longer follows Hooke’s
law, hence the curve changes from a
linear region to a nonlinear region.
• The stress corresponding to the end of
the elastic region of the curve is known
as the yield stress (σy) and the strain at
this value is the yield point strain (εyp).
However, this transition is not always
readily apparent for certain materials and
the change can be difficult to pinpoint. In
this case, it is common to use a 0.2%
strain offset to determine the yield point.
• After yielding, there is an increase in stress
required to continue plastic deformation until a
maximum is reached (the ultimate tensile
strength, σuts).
• After this value, necking of the specimen
• With the onset of necking, the stress required
to cause further plastic deformation decreases
until the point of material fracture (fracture
strength, σf).
(tensile testing: Causes of Plastic Deformation – Metals
• Metals undergo deformation due to dislocation glide along a plane called the slip plane.
• Slip occurs when there is a great enough force in an orientation that coincides with a
slip plane.
• The magnitude of the shear stress felt by the slip plane
(resolved shear stress, τr) is of particular interest and can be
determined by the equation:
φ: angle between the normal to the slip plane and the direction
of applied force
λ: angle between the slip direction and the direction of applied
• When the resolved shear stress along a slip plane exceeds a
certain value, called critical resolved shear stress (τcrss), slip is
initiated. Causes of Plastic Deformation – Amorphous Polymer and Ceramics (Glasses)
• Amorphous polymers and ceramics deform by viscous flow.
• Shear stress also plays an important role in viscous deformation.
• In viscous flow, the rate of deformation is proportional to the applied stress.
• The proportionality constant η is the viscosity of the sample, and ϒ is the rate of
shear deformation (dϒ/dt). This is called Newton’s law.
• Newton’s law can be seen as an extension of the shear stress/shear strain
relationship for solids.
As in the figure, another form of ϒ is used:
However, amorphous materials can be thought of as cooled liquids, so when we replace the
solid sample in the figure with an equal volume of liquid, we obtain a relationship like:
In this case, the shear force does not
cause a single strain value, but the
deformation continues with time. The
rate of deformation [(d(dy/dt))] is
proportional to the shear force F:
If the force is normalized by the area to obtain shear stress (τ) and the deformation rate is
normalized by the height (dx) the expression becomes
Because dx is constant, the equation can be rearranged:
Replacing dy/dx with ϒ, and including η as a constant of proportionality yields
A sample was prepared for tensile testing. The initial dimensions of the rectangular
specimen were 30 mm long and 15 mm wide, with an average thickness of 3 mm. The
mechanical testing was conducted at a rate of 5 mm/sec. The following data were
* Assume that 5mm of the specimen length is clamped by the testing grips at each end,
such that the initial gauge length of the specimen is 20mm.
Gauge length
Force (N)
Stress (Pa,
1. Calculate the engineering stresses and strains from the information given.
2. Plot the stress-strain curve and discuss the type of deformation (elastic or plastic
A sample was prepared for tensile testing. The initial dimensions of the rectangular
specimen were 30 mm long and 15 mm wide, with an average thickness of 3 mm. The
mechanical testing was conducted at a rate of 5 mm/sec. The following data were
* Assume that 5mm of the specimen length is clamped by the testing grips at each end,
such that the initial gauge length of the specimen is 20mm.
BIOE 3030: Introduction to
5. Biomedical Degradation
5.2. Corrosion of metals
• Knowledge of corrosion is crucial in predicting the biocompatibility of an
• During corrosion, leaching of ions from the metallic surface into the
surroundings may have deleterious biological consequences.
5.2.1. Fundamentals of corrosion
• Corrosion is an electrochemical process.
• Corrosion occurs through coupling of oxidation generating electrons and
reduction consuming electrons.
Electrochemical cell
Oxidation takes place at the anode.
Reduction takes place at the cathode.
Au, Cr, Al, Ag, Na
Possible combinations to make electrochemical cell. Suppose that two
electrodes have equal number of valence ions. Nernst equation
• Electrochemical potential between the electrodes.
overall reaction
Electrochemical potential in standard condition (room temp)
Electrochemical potential in non-standard condition
R: gas constant, 8.314 (J/K mol)
T: absolute temperature (K)
F: Faraday’s constant, 96,500 (coulombs/mol)
n: valence of ions
5.3. Degradation of Polymers
5.3.2. Chain Scission by Hydrolysis
• For polymers undergoing biodegradation via hydrolysis, two mechanisms can
be distinguished: bulk and surface degradation.
• In bulk degradation, when the polymer sample is exposed to water, the water
molecules infiltrate into the polymer chains and hydrolysis occurs. And the
hydrolysis creates voids.
• In surface degradation, when the polymer sample is exposed to water, the
hydrolysis occurs at the interface of the polymer and water. Degradation Mechanisms
You have these materials.
Au, Cr
1. Show oxidation and reduction reactions at anode and
2. Assuming that the concentrations of the solutions in each
compartment are 1M and the cell is tested at room
temperature, calculate the potential difference between the
electrode in each electrochemical cell.
3. Consider now that the cells are tested at body temperature.
Calculate the potential difference between the electrodes.
Assume that the concentrations of the materials at anode and
cathode are 100 and 110 µg/dL.
Oct 5
Oct 7
Oct 12
Contact Angle (2)
AFM (3)
XRD (2)
Energy Dispersive X-ray (3)
Confocal Microscopy (2)
XPS (2)
Zeta Potential (2)
We will do group presentations about Surface Characterization Techniques.
1. Park will run a software to randomly assemble 9 teams and post the teams on Canvas.
2. We will do 3 presentations/class. Prepare 15 min presentation. Presentation must contain
background of each characterization technique, any theory for the equipment, sample
preparation and measurement, data processing, and examples in scientific literatures. Since it is
team presentation, all members should present certain portion.
3. After presentation, there will be ~5min Q&A section.

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