1. Electric Circuits and Resistance

2. Learning Objectives Relate voltage, amperage and resistance with Ohm’s Law. Describe the components of a simple electric circuit. Calculate the total resistance, amperage and voltage drop for series circuits. Calculate the total resistance, amperage and voltage drop for parallel circuits.

3. Ohm’s Law The number of electrons flowing through a conductor is related to the resistance of the conductor (friction) and the force pulling the electrons (voltage). –Increase the voltage for the same resistance and the amperage increases –Increase the resistance and increased voltage is required to move the same amperage –Increase the amperage and an increased voltage is required to move against the same resistance Formula: –V = IxRI = V/RR = V/I V - VoltageI - Current (Amps) R - Resistance (Ohms) Importance: –Critically important in wiring buildings, electronics and the operation of appliances

4. Using Ohm’s Law In basic Ohm’s Law problems you are given 2 of the 3 quantities and asked to solve for the unknown. A lamp draws a current of 0.91A with a potential difference of110V. What is the resistance in the lamp? –Given:Current = 0.91 AmpsVoltage = 110V –Needed:Resistance –Formula:R = V/I –Solution:R = 110V / 0.91A = 121  The bulb in a flashlight draws 1.7A with new batteries producing a potential difference of 3.0V. What is the current when the voltage of the batteries drops to 2.5V? –Given: New Voltage = 3.0V; New Current = 1.7A; Old Voltage = 2.5V –Needed: Resistance; Old Current –Formula:R = V/I& I = V/R –Solution (Part 1):R = 3.0V/1.7A = 1.76  –Solution (Part 2):I = 2.5V / 1.76  = 1.4A

5. Ohm’s Law Problems Ohm’s Law Formulas: V = IRI = V/RR = V/I –V – Voltage (V)I – Current (Amps)R – Resistance (Ohms,  ) 1.Determine the resistance in a 6V circuit with a current of 0.5A. 2.Calculate the voltage necessary to move 3.1A of current against 200  of resistance. 3.What resistance would be necessary to produce a 3A current in a 12.8V circuit? 4.What is the resulting current if 100V of potential difference exist across a circuit with 2,100  of resistance? 5.What is the resistance in a light bulb if 0.5A of current is generated by 13.2V? 6.What is the resistance of the human body if 60V of potential difference produces 0.003A of current?

6. Basic Electric Circuits Basic electrical circuits are composed of 4 major parts. –Voltage Source - There has to be a voltage difference present to make electrons move Batteries or capacitors can produce Direct Current Alternators and generators produce Alternating Current Several batteries may be hooked together to increase the Voltage –Conductor - There has to be a connection (usually wire) between the high charge and low charge areas of the circuit Gold, Silver, Copper and Aluminum are the most common conducting materials used in electrical circuits –Switch - Working circuits must have a switch to turn the circuit on and off On/off switch Breaker or Fuse –Load - Light, motor, or other electrical appliance that turns electrical energy into some other form of energy There may be several loads in a single circuit

7. Electric Circuit Diagrams Voltage Source - Parallel short & long lines –Positive side - long; negative - short –Electricity always flows negative to positive Loads - Zig zag line - electrical appliance or light bulb –Loads are the locations where electrons actually do work –Loads often convert electricity to different forms of energy or work. –Loads produce resistance to electrical movement Switch - Arrow or line between 2 dots –May be closed (straight) or open (tilted up) Conducting Wire - Thin lines connecting other components

8. Series Circuits Consists of multiple resistors lined up one behind the other –Current has to flow through all resistors –Total circuit resistance = Sum of individual resistances Formulas: Total Resistance:R total = R 1 +R 2 +R 3 ….+R N Amperage:A 1 = A 2 = A 3 = A total Voltage Drop: V 1 = AxR 1 V 2 = AxR 2 V total = V 1 + V 2 If any resistor fails, current stops The voltage drops a fraction of the total circuit voltage for each resistor

9. Example Problem A circuit containing a 30 , a 5 , and a 50  resistor in series has a potential difference of 60V. Determine the total resistance, circuit amperage and voltage drop across each resistor. Solution: R total = R 1 + R 2 + R 3 = 30  + 5  + 50  = 85  I = V/R total = 60V/85  = 0.71A V 1 = IR 1 = 0.71A x 30  = 21.2V V 2 = IR 2 = 0.71A x 5  = 3.55V V 3 = IR 3 = 0.71A x 50  = 35.5V V total = V 1 + V 2 + V 3 = 21.2V + 3.55V + 35.5V = 60.25 (Close enough!)

10. Parallel Circuits Made up of several resistors set side by side. Current splits going through each resistor simultaneously. Formulas: Resistance:1/R total = 1/R 1 + 1/R 2 + 1/R 3 +…. Amperage:I 1 = V/R 1 I 2 = V/R 2 I total = V/R total Voltage:V 1 = V 2 = V 3 = V total If any resistor fails the current just travels through the other resistors

11. Solving Total Resistance in Parallel Circuits Formula:1/R total = 1/R 1 + 1/R 2 + ….. Solution Process: –Write the formula containing the individual resistances –Find a common denominator for the individual resistors –Add the resistance fractions –Invert answer to get total resistance Example Problem: Find R total for a parallel circuit containing a 5  and a 40  resistor. Solution: –Write Formula: 1/R total = 1/5 + 1/40 –Find common Denominator:1/R total = 8/40 + 1/40 –Add Resistance Fractions:1/R total = 9/40 –Invert Answer:R total = 40/9 = 4.44 

12. Example Problem A circuit containing a 30 , a 5 , and a 50  resistor in parallel has a potential difference of 60V. Determine the total resistance and circuit amperage. Solution: R total = 1/R 1 + 1/R 2 + 1/R 3 = 1/30  + 1/5  + 1/50  = 1/.253  = 3.9  I = V/R total = 60V/3.9  = 15.38A

13. More Resistance in Parallel Circuits Calculate the total resistance in a circuit containing a 30 , 25 , 50 , and 20  resistor in parallel. Formula: Solution: 1/R total = 1/R 1 + 1/R 2 + 1/R 3 + 1/R 4 1/R total = 1/30 + 1/25 + 1/50 + 1/20 Common Denominator - 300 1/R total = 10/300 + 12/300 + 6/300 + 15/300 = 43/300 R total = 300/43 = 6.98 

14. Problems Draw diagrams for the circuits described below and calculate R Total and I Total 1.A 60V circuit with a 25 , 10  and 100  resistor hooked in series. 2.A circuit where 2-40  resistors are hooked in parallel to a 12V power source. 3.A circuit composed of a 36V power supply connected to 4-5  resistors in series. 4.A 24V power source connected to 2-10  and 3-5  resistors hooked in series. 5.A circuit composed of 4 parallel 100  resistors hooked in series to a 200V power supply.

15. Combination Circuits In order to create circuits with odd resistances you often have to use a combination of series and parallel resistors The circuit overall is treated like a series circuit, with any parallel resistors combined as if they were a single “equivalent” resistors –Ex. 2- 500  parallel resistors would act like a single 250  resistor 1-500 + 1/500 = 2/500 500/2 = 250  –Combine all parallel combinations then add the resistances

16. Combination Series & Parallel Circuits Circuits that have series and parallel resistors are solved as follows: –Calculate total resistance only on parallel resistors first –Treat the total for the parallel resistors as a single resistor, and calculate total resistance for it and the other series resistors Example Problem: –A circuit contains 2 -50  and 1-40  resistors in parallel, followed by a 60  and a 30  resistor in series. Calculate the total resistance. Solution: –Parallel Resistors: R total = 1/50 + 1/40 = 4/200 + 5/200 = 9/200 = 200/9 = 22.2  –Series Resistors: R total = 22.2  + 60  + 30  = 112.2 

17. Practice Problem Determine the following for the circuit diagrammed below: a) Total Circuit Resistance b) Total Circuit Amperage c) Voltage drop across the 25  and 5  resistors 25  40  10  55 50V

18. Practice Problem Solution Combine all parallel resistors:1/R = 1/40 + 1/10 = 5/40  40/5 = 8  Combine all series resistors:R Total = 25 + 8 + 5 = 38  Divide voltage by R Total :I = 50/38 = 1.32A Multiply R 1 and I:V 1 = 25x1.32 = 33V Multiply R 3 and I:V 3 = 5x1.32 = 6.6 25  40  10  55 50V Series Parallel