SVGS Physics

A resistor provides resistance, (R), measured in Ohms (Ω). Resistance slows the current through a circuit.

A battery or power supply provides voltage/potential difference/electric potentia(V) measured in Volts (V)

The circuit’s current (I) is the amount of charge per second passing through a section of wire.

Units of current = Coulomb/second = Ampere (A)

V = IR

Can be applied to the entire circuit The overall voltage is equal to the total current times the

equivalent resistance

Can be applied to a single resistor The voltage drop across the resistor is equal to the

current passing through it times its resistance

Can be applied to a simplified group of resistors within a circuit

Through the course of one path around the circuit, the voltage gained = the voltage dropped.

If the circuit has multiple path options, EACH path results in the same voltage drop (Kirchoff’s Voltage Rule)

Series Parallel

Image: Wikipedia. Accessed 2/1/15 from http://en.wikipedia.org/wiki/Series_and_parallel_circuits

Image: Physics 24/7. Accessed 2/1/15 from http://www.physics247.com/physics-

tutorial/parallel-circuits.shtml

Series Circuit Req = R1 + R2 + R3 + …

Parallel Circuit (1/Req)= (1/R1) + (1/R2) + (1/R3) + …

Image: Scontent. Accessed 2/16/14 from https://scontent-b.xx.fbcdn.net/hphotos-

prn2/t1/1656382_740686815950311_880193648_n.jpg

The total current entering a junction equals the total current leaving a junction. In this image: I1 = I2 + I3

Image: Brigham Young University. Accessed 2/16/14 from http://www.physics.byu.edu/faculty/christensen/Physics%20220/FTI/28%20DC

%20Circuits/28.11%20Kirchoff's%20Junction%20Rule.htm

Remember from the voltage lecture that “voltage” is simply the change in potential energy divided by one of the charges?

Therefore,

ΔU = qV

In PHYS 140, you learned that power was the change in energy over time.

Therefore,

However, we already learned that ,

so

So, the main takeaway: (and the units are still Watts)

P =ΔUt

ΔU = qVP =

qVt

=qtV

P = IV

A flashlight is powered by two 1.50 V batteries in series. The batteries are connected to the light bulb by wires of total resistance 0.40 Ω. The light bulb itself has a resistance of 9.70 Ω. What is the power supplied by the batteries?

Image: Wikipedia. Accessed 2/17/14 from http://en.wikipedia.org/wiki/Maglite

Knob to switch between voltage, current, resistance, and capacitance

Must be wired differently for different measurements

Image: Electrical Construction and Maintenance. Accessed 2/17/14 from

http://ecmweb.com/basics/how-use-digital-multimeters

Ammeter Voltmeter

Measures current

Placed in series with point of interest

Very low resistance

Measures voltage

Placed in parallel with point of interest

Very high resistance

http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc-virtual-lab

Follow the instructions provided to analyze the battery and bulb circuitsyou have already become familiarwith!

Image: Home Tools. Accessed 2/17/14 from http://hometools.onsugar.com/Digital-Multimeter-13149930

The voltage across the battery and the bulb were not exactly the same!

The measured resistance of the bulb was way lower than the calculated resistance given the voltage and current measurement!

Why??Image: Know Your Meme. Accessed 2/17/14 from

http://knowyourmeme.com/photos/96044-trollface-coolface-

problem

Proportionality between the resistance of a conductor and its properties

For a wire, these properties include the length of wire and the cross-sectional area:

L is the length, A is the cross-sectional area, R is the resistance of the wire, and ρ is the resistivity (units = Ω m)

R = ρLA

The hotter the wire, the more resistivity:

where ρ0 is a known resistivity (generally at 20°C), α is the temperature coefficient of resistivity, and ΔT is the change in temperature from the known value (i.e., change from 20°C).

ρ = ρ0 1+αΔT( )

Every element has a different resistivity

For wires, the lower the resistivity, the more “ideal” the wire is

Resistivity chart: Pg. 646 in textbook

J. Random Troll uses an extension cord to power his outdoor Super Bowl viewing arena. It is made of two copper wires of diameter 0.912 mm (one carrying current to the arena, one carrying current from the arena). The wire is currently at 20°C and has a length of 30.0 m. What is the resistance of each wire?

After being plugged in for an extended period of time, the same wires heat to 100°C. What is the resistance in each wire now?

Image: Portable Fireplace. Accessed 2/17/14 from

http://www.portablefireplace.com/blog/extension-cords-electric-fireplaces/

where Q is the charge on each plate, ΔV is the voltage difference between the plates, and C is the capacitance

Units of capacitance = Coulombs/Volt = farads (F)

Q = C Δ V

Image: Busse. Accessed 3/15/13 from http://whs.wsd.wednet.edu/faculty/busse/mathhomepage/busseclasses/apphysics/studyguides/chapter15

16/electricfieldlinespart2.html

When switch is open, no current is flowingThe instant the switch is closed, the uncharged capacitor acts as a wire (short circuit)After a while, thecapacitor builds up charge to reach a voltage equal to thebattery’sNo current flows(break in wire) Battery

Switch

Capacitor

Work together as batteries and capacitor plates to complete a full circuit and learn about all the wonders that abound!

Capacitors in series have the same magnitude charge on each plate. Why? Additionally, the total potential difference across all capacitorsmust still be the same as the EMF.

Image: Stony Brook University. Accessed 2/20/14 from

http://www.ic.sunysb.edu/Class/phy141md/doku.php?id=phy142:lectures:8s13

Because , and all capacitors have the same charge Q, solve for the change in voltage. This sum must be the same as the EMF. Therefore,

If you divide all arguments by Q, you’ll find the main takeaway – that equivalent capacitance in series is:

Q = C Δ V

...321

+++=CQ

CQ

CQ

CQ

eq

...1111

321

+++=CCCCeq

Each capacitor has the same change in voltage across its plates, but not necessarily the same charge

Image: Rapid Tables. Accessed 2/20/14 from http://www.rapidtables.com/electric/capacitor.htm

Again, considerIf a total charge Q is pumped out of a battery toward a parallel capacitor circuit, then Q = q1 + q2 + q3 +…Therefore,Divide by V, and you’ll find the main takeaway for parallel capacitors:

Q = C Δ V

...321 +++=⋅ VCVCVCVCeq

...321 +++= CCCCeq

Assume C1 = 2μF, C2 = 3μF, and C3 = 6μF and V = 12 V. A) What is the equivalent capacitance? B) What is the magnitude of charge on each plate? C) What is the voltage stored in each capacitor?

Image: Stony Brook University. Accessed 2/20/14 from

http://www.ic.sunysb.edu/Class/phy141md/doku.php?id=phy142:lectures:8s13

Assume C1 = 2μF, C2 = 3μF, and C3 = 6μF and V = 12 V. A) What is the equivalent capacitance? B) What is the magnitude of charge on each plate? C) What is the voltage stored in each capacitor?

Image: Rapid Tables. Accessed 2/20/14 from

http://www.rapidtables.com/electric/capacitor.htm

Circuit containing both capacitors and resistorsOften used for timersExamples: windshield wipers, strobe lights, pacemakersWhen R and C are in series, at any given time,ε = VR + VCwhere ε is the voltage of the battery.

Image: Sinclair Community College. Accessed 2/20/14 from http://people.sinclair.edu/nickreeder/eet155/mod02.htm

Switch is initially open and capacitor is unchargedThe instant the switch is closed, all voltage drop is over the resistor (the uncharged capacitor is essentially a wire).After the switch has been closed for a while, the capacitor holds the same voltage as the battery, and almost no current flowsAt this point, the capacitor is essentially a “break” in the circuit.

With both a resistor and a capacitor hanging out in a circuit, there’s a lot of units to deal with – Farads, Ohms, Amperes, Volts, Coulombs…To simplify things, find and simplify the units of (resistance) x (capacitance) or RC. (Hint: It’s a base SI unit).

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http://www.nicheadsensechallenge.com/niche-site-challenge-overview/

The unit for R is Ohm, which is equivalent to Volts/AmpThe units for C is Farad, which is equivalent to Coulomb/Volt. So, RC has a unit of Coulomb/Amp. However, Amp is a Coulomb/secondTherefore, RC has a unit of Coulomb/Coulomb/secondThe unit for RC is simply a second!

If RC is a unit of time, let’s see how the capacitor’s voltage is measured on a voltage vs. time graph!

What sort of function does this represent?

Image: University of New Mexico. Accessed 2/24/14 from

http://www.unm.edu/~toolson/rc_circuit.html

While charging a capacitor, you can find the voltage stored at any time through the following equation:

constant timeRC theis whereττ=RC

( )τε /1 tC eV −−=

Voltage vs. Time graph

Image: Planet Arduino. Accessed 2/24/14 from

http://www.planetarduino.org/?cat=357

Given the equation for the voltage across the capacitor at any given time while it is charging in a simple series RC circuit, derive an equation for the voltage across the resistor.

Image: Litchicblog. Accessed 2/23/14 from http://litchicblog.wordpress.com/2014/01/21/challenge-

accepted/

Knowing that I = V/R, and given the expression you derived for voltage, the current through an RC circuit as a capacitor charges is:

τε /teR

I −=

Two 0.5 μF capacitors are connected in series to a 50 V battery through a 4 MΩ resistor at t = 0. The capacitors are initially uncharged.

A) Find the charge on the capacitors at t = 1s and at t = 3s.

B) Find the current through the circuit at the same two times

Image: University of New Mexico. Accessed 2/24/14 from

http://www.unm.edu/~toolson/rc_circuit.html

A charged capacitor (at max voltage) is hooked up to a resistor

The capacitor then releases charge and its voltage decreases exponentially:

τε /tC eV −⋅=

Image: Interface Bus. Accessed 2/27/14 from http://www.interfacebus.com/Design_RC_Falling_Time_Co

nstant_Graph.html