SMV ELECTRIC TUTORIALS Nicolo Maganzini, Geronimo Fiilippini, Aditya Kuroodi 2015 Relevant Course(s): EE10, EE11L

RESISTORS IN NETWORKS 2

What are we learning?  Learn about the math behind networks of resistors.  Current and Voltage laws.  Predicting/designing circuits that have specific values of  Current, Voltage, Resistance  Learn about some very important structures of networks  Parallel and series  How are they used?  CAUTION: Math involved. 3

Resistors in Networks 4  In Circuit Schematics:  In Real Life:

Resistor Network Calculations - Series Networks  You have this circuit: R1 = 1 Ohm, R2 = 2 Ohm, R3 = 3 Ohm, V = 6V  How can you apply Ohm’s law to find out how much current is flowing? 5

Series Resistors Equation.  This is called a series connection:  Equivalent Resistance = R1 + R2 + R3 + R4  Since there is only one path for electrons, there is only one current value in the part of the circuit with the series connection.  Try it yourselves! (next slide) 6

The circuit we’re building:  R1 = 100 Ohm  R2 = 220 Ohms  R3 = 300 Ohms  Battery = 9V  Measure current at nodes 1,2. Write them down. Check that they are equal.  Measure voltages V1(across R1), V2 (across R2), V3 (across R3), across the battery.  Calculate:  V1/R1, V2/R2, V3/R3  What should these be equal to?  V1+V2+V3  What should this be equal to?  (V1+V2+V3)/(R1+R2+R3)  What should this be equal to? 7

Parallel Networks  Current has multiple paths it can take.  It will split according to the resistance in each path.  Path with lower resistance gets most current.  Path with higher resistance gets less current.  If resistances are equal, all paths have the same current. 8

Let’s combine the two!  Split circuit between parallel and series parts.  Simplify the parallel part and add it to the series part.  Parallel part simplification:  Overall equation for resistance: 9 This is in Parallel: Find it’s equivalent Then add it to this one!

Sample Problem  Calculate the current flowing out of the battery in this circuit:  R1 = 100 Ohms  R2 = 150 Ohms  R3 = 200 Ohms  Battery = 9V 10

Kirchoff current and voltage laws  How do we analyze more complicated circuits?  There are some physics laws that we can apply to circuits that allows us to find equations: Kirchoff laws.  Steps:  1) Apply Laws  2) Find Equations  3) Solve equations to find current, voltage and resistance. 11

Kirchoff Voltage Law (KVL)  What the law says:  The sum of all voltages in a loop must be equal to zero.  Example of how we use it:  Vbatt = 9V.  V1 = 2V  V2 = 3V  R3 = 4 Ohms  Find the current in the circuit. 12

 Step 1) Apply law:  The voltage produced by the battery is equal to the voltage dropped by each resistor.  Step 2) Find Equation:  Vbatt = V1+V2+V3  Know Vbatt, V1, V2; Find V3  I = V3/R3  Know V3 and R3, Find I.  Step 3) Solve:  V3 = = 4V  I = 4/4 = 1A 13 Kirchoff Voltage Law (KVL)

Kirchoff Current Law (KCL)  What the law says:  The sum of all currents entering and exiting a node must be zero.  Example of how we use it:  R1 = 100 Ohms.  R2 = 200 Ohms  R3 = 200 Ohms.  Current through R1 = 1A  Find voltage of battery. 14

 Step 1) Apply Laws:  Current flowing into node 2 from R2 and R3 must be equal to current flowing out towards R1.  Current flowing in R2 and R3 must be equal because resistances are equal (200 ohm)  Sum of voltages must be equal to the battery voltage  Step 2) equations:  I1 = I2 + I3  I2 = I3  V1+V2 = V1 + V3 = Vbattery  Step 3) solve:  1 = ½ + ½  I2 = I3 = ½ A  V1 = I1 R1 = 100V  V2 = V3 = ½ x 200 = 100V  Vbatt = = 200V 15 Kirchoff Current Law (KCL)

Using series connections to make a sensor 16

CAPACITORS AND SIGNAL FILTERING

Ohm’s Law for Capacitors 18

Low Pass Filter

High Pass Filter 20 R = 1 kOhm C = 0.22 microF

What have we learned?  If the signal has a certain frequency, we can make an R-C circuit that cancels the signal out.  If a signal has more than one frequency, such as noise:  Can clean it up using an R-C filter designed to cancel out all frequencies lower than a certain amount. 21