 Current Electricity, Ohm’s Law & Circuits. Current (I) The rate of flow of charges through a conductor Needs a complete closed conducting path to flow.

Presentation on theme: "Current Electricity, Ohm’s Law & Circuits. Current (I) The rate of flow of charges through a conductor Needs a complete closed conducting path to flow."— Presentation transcript:

Current Electricity, Ohm’s Law & Circuits

Current (I) The rate of flow of charges through a conductor Needs a complete closed conducting path to flow Must have a potential difference (voltage) Measured with an “ammeter” in amps (A) named for Ampere – French scientist I = current, A Q = charge, C t = time, s So: 1 Amp = 1

Voltage (V) Electric potential difference between 2 points on a conductor. Equal to the electric potential energy per charge. Sometimes described as “electric pressure” that makes current flow Supplies the energy of the circuit Measured in Volts (V) using a voltmeter 1 Volt = 1 Joule / Coulomb

Resistance (R) The “electrical friction” encountered by the charges moving through a material. Depends on material, length, and cross- sectional area of conductor Measured in Ohms (Ω) Where: R = resistance, = length of conductor, A = cross-sectional area of conductor, ρ = resistivity of conducting material

Resistivity ( ρ ) Property of material that resists the flow of charges (resistivity, ρ, in Ωm) The inverse property of conductivity Resistivity is temperature dependent…as temperature increases, then resistivity increases, and so resistance increases.

Ohm’s Law A relationship between voltage, current, and resistance in an electric circuit used to make calculations in all circuit problems V = potential difference (voltage) in volts I = electric current in amperes (amps, A) R = resistance in ohms (  )

Electric Power (Watts) Used for thermal energy

Electric Energy Electric energy can be measured in Joules (J) or Kilowatt hours ( kWh ) for Joules use Power in watts and time in seconds for kWh use Power in kilowatts and time in hours

Series Circuits Current can only travel through one path Current is the same through all parts of the circuit. The sum of the voltages of each component of the circuit must equal the battery. The equivalent resistance of a series circuit is the sum of the individual resistances. R1R1 R2R2 R3R3 V I

Solving a Series Circuit 6V R 1 =1 Ω R 2 =1 Ω ITIT Step 1: Find the equivalent (total) resistance of the circuit Step 2: Find the total current supplied by the battery Step 3: Find Voltage Drop across each resistor. Note: Since both resistors are the same, they use the same voltage. Voltage adds in series and voltage drops should add to the battery voltage, 3V+3V=6V

Parallel Circuits Current splits into “branches” so there is more than one path that current can take Voltage is the same across each branch Currents in each branch add to equal the total current through the battery R1R1 R2R2 R3R3 V

Solving a Parallel Circuit R 1 =1 Ω R 2 =2 Ω R 3 =3 Ω 12V Step 1: Find the total resistance of the circuit. Step 2: Find the total current from the battery. Step 3: Find the current through each resistor. Remember, voltage is the same on each branch. Step 4: Check currents to see if the answers follow the pattern for current. The total of the branches should be equal to the sum of the individual branches.

Combo Circuits with Ohm’s Law What’s in series and what is in parallel? 15V 3Ω3Ω 5Ω5Ω 7Ω7Ω 1Ω1Ω 2Ω2Ω 6Ω6Ω 4Ω4Ω It is often easier to answer this question if we redraw the circuit. Let’s label the junctions (where current splits or comes together) as reference points. AB CD 3Ω3Ω A B 6Ω6Ω 4Ω4Ω 1Ω1Ω C 5Ω5Ω 2Ω2Ω D 7Ω7Ω 15V

Combo Circuits with Ohm’s Law Now…again…what’s in series and what’s in parallel? 3Ω3Ω A B 6Ω6Ω 4Ω4Ω 1Ω1Ω C 5Ω5Ω 2Ω2Ω D 7Ω7Ω 15V The 6Ω and the 4Ω resistors are in series with each other, the branch they are on is parallel to the 1Ω resistor. The parallel branches between B & C are in series with the 2Ω resistor. The 5Ω resistor is on a branch that is parallel with the BC parallel group and its series 2Ω buddy. The total resistance between A & D is in series with the 3Ω and the 7Ω resistors.

Combo Circuits with Ohm’s Law Finding total (equivalent) resistance 3Ω3Ω A B 6Ω6Ω 4Ω4Ω 1Ω1Ω C 5Ω5Ω 2Ω2Ω D 7Ω7Ω 15V To find R T work from the inside out. Start with the 6+4 = 10Ω series branch. So, 10Ω is in parallel with 1Ω between B&C… Then, R BC + 2Ω=2.91Ω and this value is in parallel with the 5Ω branch, so… Finally R T = R AD +3 + 7 = 1.84 + 3 + 7 R T = 11.84Ω

Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor 3Ω3Ω A B 6Ω6Ω 4Ω4Ω 1Ω1Ω C 5Ω5Ω 2Ω2Ω D 7Ω7Ω 15V R T = 11.84Ω I T =1.27A The total current I T goes through the 3Ω and the 7Ω and since those are in series, they must get their chunk of the 15V input before we can know how much is left for the parallel. So… Then… So… Since parallel branches have the same current, that means the voltage across the 5Ω resistor V 5Ω =4.84V and the voltage across the parallel section between B&C plus the 2Ω is also 4.84V

Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor (continued) Known values from previous slide. To calculate the current through the 5Ω resistor… To calculate the top branch of the parallel circuit between points A & D we need to find the current and voltage for the series 2 Ω resistor. Since the current through the resistor plus the 0.92A for the bottom branch must equal 1.3A. So… 3Ω3Ω A B 6Ω6Ω 4Ω4Ω 1Ω1Ω C 5Ω5Ω 2Ω2Ω D 7Ω7Ω 15V I T =1.27A I 2Ω =0.81A I 5Ω =0.46A

Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor (continued) Known values from previous slide. 3Ω3Ω A B 6Ω6Ω 4Ω4Ω 1Ω1Ω C 5Ω5Ω 2Ω2Ω D 7Ω7Ω 15V I T =1.27A I 2Ω =0.81A I 5Ω =0.46A Next we need to calculate quantities for the parallel bunch between points B&C. The voltage that is left to operate this parallel bunch is the voltage for the 5Ω minus what is used by the series 2Ω resistor. The 1Ω resistor gets all of this voltage. Finally we need to calculate the current through the 6Ω and 4Ω resistors and the voltage used by each. All we need now is the voltage drop across the 6Ω and 4Ω resistors. So… I 1Ω =0.68A I 6Ω =I 4Ω =0.068A THE END!

Voltmeter and Ammeter Ammeter –Measures current in amps –Placed in series where current is to be measured Voltmeter –Measures voltage in Volts –Placed in parallel across whatever is being measured

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