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Current, Resistance,Voltage Electric Power & Energy Series, Parallel & Combo Circuits with Ohm’s Law, Combo Circuits with Kirchoff’s Laws Current (I) • The rate of flow of charges through a conductor • Needs a complete closed conducting path and a potential difference or voltage to flow • Measured with an “ammeter” in amps (A) named for Ampere – French scientist • I = current, A q = charge, C Δt = time,s (Note: 1 A = 1 C/s) q I t Voltage (V) • Electric potential difference between 2 points on a conductor • Sometimes described as “electric pressure” that makes current flow • Supplies the energy of the circuit • Measured in Volts (V) using a voltmeter Resistance (R) • Opposition to the flow of charges. “electrical friction” of the moving charges. • Measured in Ohms (Ω) • Resistance in a wire depends on - material (low R for copper) - temperature (as T increases, R increases) - length (as length increases, R increases) - cross-sectional area ( as A incr., R decr.) Resistance (R) • The “electrical friction” encountered by the charges moving through a material. • Depends on material, length, and crosssectional area of conductor • Measured in Ohms (Ω) R A Where: R = resistance, = length of conductor, A = cross-sectional area of conductor, ρ = resistivity of conducting material Note: Don’t need to know this formula. 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 ( ) V I R V IR Voltmeter and Ammeter • Ammeter – measures current in amps or mA – must be placed in series with what is being measured • Voltmeter – measures voltage in volts – must be placed in parallel with what is being measured Electric Power (Watts) Energy Power time 2 V P IV I R • Used for thermal energy R 2 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 E Pt 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. Req R1 R2 R3 ... VT V1 V2 V3 ... VBattery IT I1 I 2 I 3 ... R1 V I R3 R2 Solving a Series Circuit Step 1: Find the equivalent (total) resistance of the circuit R1=1 Ω RT 1 2 3 IT 6V RT R1 R2 R2=2 Ω Remember: The current is the same throughout a series circuit so… IT = I1 = I2 Step 2: Find the total current supplied by the battery VT 6V IT 2A RT 3 Step 3: Find Voltage Drop across each resistor. V1 I1 R1 2 A 1 2V V2 I 2 R2 2 A 2 4V 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 1 1 1 1 ... Req R1 R2 R3 I T I1 I 2 I 3 ... V VBattery V1 V2 V3 ... R1 R2 R3 Solving a Parallel Circuit R2=2Ω 12V R3=4Ω R1=1Ω Step 3: Finding the current through each resistor. Remember, voltage is the same on each branch as the battery voltage. Note: Step 1 & 2 were not necessary to do step 3!! I1 Step 1: Finding the total resistance of the circuit. 1 R1 1 R2 RT 1 1 1 2 RT 1 R3 1 1.75 so... RT 0.572 Step 2: Finding the total current from the battery. VT 12V T RT .572 I 21A 1 VT R1 121V 12 A I 2 VR22 VRT2 122 V 6 A I3 1 1 4 V1 R1 V3 R3 VRT3 124 V 3 A Step 4: Check currents to see if the answers follow the pattern for current. I T I1 I 2 I 3 I T 12 A 6 A 3 A 21A The total of the branches should be equal to the sum of the individual branches. Solving Combo Circuits – Find RT first R3 = 2 Ω R2 = 7 Ω R1 = 4 Ω R4 = 3 Ω V = 15 V Step 1: Find RS for any series within a R2-3 = 9 Ω R1 = 4 Ω R4 = 3 Ω parallel branch. R2-3= 7 + 2 = 9Ω V = 15 V Step 2: Find RP for any parallel parts. 1 R2 4 1 1 RP 2.25 9 3 Step 3: Find RT for the entire circuit. RT 2.25 4 6.25 R1 = 4 Ω R2-4 = 2.25 Ω V = 15 V R1-4 = 6.25 Ω V = 15 V Solving Combo Circuits VT 15V 2.4 A RT 6.25 Step 4: Find IT using Ohm’s law. IT Step 5: Find the voltage across any resistors in series with the battery. V1 I1R1 IT R1 Step 6: Find the voltage drop across the whole parallel part. (2 methods are shown.) V1 2.4 A 4 9.6V VT V1 V2 4 VP V2 4 15V 9.6V 5.4V .....or..... VP I P RP VP 2.4 A 2.25 5.4V Step 7: The voltage drop across each parallel branch is the same as the total voltage drop across the parallel part. 𝑉𝑃 = 𝑉4 = 𝑉2−3 Solving Combo Circuits Step 8: Find 𝐼4 using Ohm’s Law. Step 9: Find 𝐼2,3 using Ohm’s Law. In series, the current is same. 𝐼4 = 𝐼2,3 = 𝑉4 𝑅4 𝑉2,3 𝑅2,3 = 5.4 𝑉 3𝛀 = 1.8 A = 5.4 𝑉 9𝛀 = 0.6 A 𝐼2 = 𝐼3 = 𝐼2,3 = 0.6 A {Check: 𝐼𝑇 = 𝐼4 + 𝐼2,3 =1.8 + 0.6 = 2.4 A} Step 10: Find 𝑉2 and 𝑉3 using Ohm’s Law. {Check: 𝑉𝑃 = 𝑉2 + 𝑉3 = 4.2+ 1.2 = 5.4 V} 𝑉2 = 𝐼2 𝑅2 = 0.6A (7𝛀) = 4.2 V 𝑉3 = 𝐼3 𝑅3 = 0.6A (2 𝛀) = 1.2V More Combo Circuits with Ohm’s Law What’s in series and what is in parallel? A 3Ω B 5Ω 15V 1Ω 6Ω 4Ω 7Ω D 2Ω C 6Ω 4Ω B 1Ω 3Ω 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. A 5Ω 15V C 2Ω D 7Ω Combo Circuits with Ohm’s Law Now…again…what’s in series and what’s in parallel? 6Ω 4Ω B 3Ω 1Ω A C 2Ω D 7Ω 5Ω 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 & D 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 6Ω 4Ω B 3Ω C 2Ω 1Ω A D 7Ω 5Ω 15V To find RT work from the inside out. Start with the 6+4 = 10Ω series branch. So, 10Ω is in parallel with 1Ω between B&C… 1 RBC 11 101 11 10 so... RBC 10 11 0.91 Then, RBC + 2Ω=2.91Ω and this value is in parallel with the 5Ω branch, so… 1 1 1 RAD 2.91 5 so... RAD 1.84 Finally RT = RAD +3 + 7 = 1.84 + 3 + 7 RT = 11.84Ω Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor RT = 11.84Ω 6Ω 4Ω IT C 1115.84V 1.27 A 2Ω B 1Ω 3Ω VT RT D A 7Ω 5Ω IT=1.27A IT=1.27A 15V The total current IT 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… IT I 3 I 7 1.27 A Then… V3 I 3 R 1.27 A 3 3.81V V7 I 7 R 1.27 A 7 8.89V So… VP 15V 3.81V 8.89V 2.3V AD Since parallel branches have the same current, that means the voltage across the 5Ω resistor V5Ω=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) 6Ω 4Ω B 1Ω 3Ω A C 2Ω I2Ω=0.81A D 7Ω 5Ω IT=1.27A Known values from previous slide. RT 11.84 I T 1.27 A V3 3.81V V5 8.89V VPAD 2.3V 15V To calculate the current through the 5Ω resistor… I 5 VR5 25.3V 0.46 A IT=1.27A 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… I 2 1.27 A 0.46 A 0.81A V2 I 2 R 0.81A 2 1.62V Combo Circuits with Ohm’s Law Solving for current and voltage drops in each resistor (continued) I6Ω=I4Ω =0.068A C B 6Ω I1Ω=0.68A A 2Ω 4Ω 1Ω 3Ω Known values from previous slide. I2Ω=0.81A D RT 11.84 7Ω V3 3.81V 5Ω IT=1.27A IT=1.27A 15V 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. I T 1.27 A Finally we need to calculate the current through the 6Ω and 4Ω resistors and the voltage used by each. I 6 I 4 0.68V ( 6 4) 0.068 A All we need now is the voltage drop across the 6Ω and 4Ω resistors. So… V7 8.89V VPAD V5 2.3V I 5 0.46 A I 2 0.81A V2 1.62V VPBC V1 0.68V I1 0.68 A VPBC V1 2.3V 1.62V 0.68V V6 I 6 R 0.068 A 6 0.41V I1 V1 R 0.68V 1 0.68 A V4 I 4 R 0.068 A 4 0.27V THE END! Kirchoff’s Laws Law of Loops ( or Voltages) treats complex circuits as if they were several series circuits stuck together. So…the rules of series circuit voltages allows us to write equations and solve the circuit. V 0 or ΣVinput = ΣVdrops Law of Nodes (or Currents) The total of the currents that enter a junction (or node) must be equal to the total of the currents that come out of the junction (or node). I 0 or ΣIin = ΣIout We use this law already in general when we add currents in the branches of a parallel circuit to get the total before it split into the branches. Kirchoff’s Law of Nodes 3Ω IT I1 I2 2Ω 2Ω 10V IT 1Ω I2 1Ω Node IT=I1+I2 The current entering one node is equal to the sum of the currents coming out Kirchoff’s Laws of Voltage writing the equations Draw current loops so that at least one Use Σ Vinput = Σ Vdrops loop passes through each resistor. for each current loop to Current loops must NOT have branches. R1 R4 R2 V IA IB IC R5 R6 R7 R3 write these equations. Remember that current is a vector so if multiple currents pass through a resistor, the total is the vector sum of the currents assuming the current loop you are writing the equation for is positive. Loop A V= IA R1+( IA- IB) R2+ IA R7 Loop B 0 = (IB- IA) R2+( IB- IC)R4+ IB R3 Loop C 0 = (IC- IB) R4+ ICR5+ IC R6 Kirchoff’s Law of Voltage putting numbers in the equations 3Ω 1. Draw current loops so that at least one loop passes through each resistor. Current loops must NOT have branches. 2. Write an equation for each loop. 3. Solve the system of equations for all of the unknowns using a matrix (next slide) 5Ω 15V IA 1Ω IB IC 6Ω 4Ω 7Ω 2Ω Loop A 15V = IA (3Ω)+( IA- IB) (5Ω) + IA (7Ω) 15 = 3IA+5IA-5IB+7IA 15 = 15 IA - 5 IB + 0 IC Loop B 0 = (IB- IA) (5Ω) +( IB- IC)(1Ω) + IB (2Ω) 0 = 5IB – 5IA +1IB -1IC+2IB 0 = -5 IA + 9 IB - 1 IC Loop C 0 = (IC- IB) (1Ω) + IC (6Ω) + IC (4Ω) 0 = 1IC-1IB+6IC +4IC 0 = 0 IA -1 IB + 11 IC Note: you must have coefficients for each unknown (even if it is zero) in every current loop equation. Kirchoff’s Law of Voltage setting up and solving the matrix for IA, IB, and IC 3Ω Beginning with the system of equations we wrote on the previous slide, we need to express these in matrix form to solve 15V for the 3 unknowns 15 = 15 IA - 5 IB + 0 IC 0 = -5 IA + 9 IB - 1 IC 0 = 0 IA -1 IB +11 IC IA 15 15 -5 0 -5 9 -1 * IB = 0 IC 0 0 -1 11 coefficients A unkowns answers * x = B Create matrix A and B in your calculator. (Matrx> >Edit, then choose A or B ) 5Ω IA 1Ω IB IC 6Ω 4Ω 7Ω 2Ω In a normal algebra equation Ax=B, the solution is x = B/A, however matrix operations do not allow for division so instead, after you create the matrices, you will use them in the following operation. x=A-1B. The answer will be in matrix form containing all of the unknowns in the order they were set up. Kirchoff’s Laws of Voltage Interpreting the answers to the matrix problem 3Ω IA 15 15 -5 0 -5 9 -1 * IB = 0 IC 0 0 -1 11 coefficients A 5Ω 15V x IC = B After performing the operation x=A-1B, the calculator will give you a matrix answer (the number of decimal places will depend on the calculator settings) like below. IC IB 6Ω unkowns answers * 1.23 IA 0.69 = I B 0.063 IA 1Ω So now we know that IA = 1.23A, IB=0.69A and IC = 0.063A Now what? 4Ω 7Ω 2Ω Using these current loop values we can now evaluate current, voltage, and power through any resistor in the circuit. Example: for the 3Ω resistor, only IA passes through it so the I3Ω = 1.23 A, the voltage is V=IR=1.23A*3Ω=3.69V, and power, P=I2R= (1.23)2*3Ω = 4.54 W Kirchoff’s Laws of Voltage But what if the resistor you ask me about is shared by two current loops? Yikes! 3Ω So now we IA 1.23 know that 0.69 = IB IA = 1.23A, 5Ω 1Ω 15V 0.063 IB=0.69A 6Ω IC I I A B IC and IC = 0.063A 4Ω 7Ω 2Ω Let’s evaluate the 5Ω resistor: Since it is shared by current loops A and B, the current is the vector sum of the two. In this case IA & IB pass through the resistor in opposite directions so…I5Ω = IA-IB=1.23A-0.69A=0.54A . The voltage drop is calculated V5Ω=I5ΩR=0.54A*5Ω=2.7V. The power dissipated is P=I5Ω2*R=(0.54A)2*5Ω=1.46 W.