2 Characteristics of Series Circuits Only one path (electron has no choices, must go through all components)If one goes out, they all go outVoltage adds VT=V1+V2+V3+….Current is constant IT=I1=I2=I3=….Resistance adds RT=R1+R2+R3+….Power adds PT=P1+P2+P3+….
3 Characteristics of Parallel Circuits More than one pathVoltage is constant VT=V1=V2=V3=….Current adds IT=I1+I2+I3+….Resistance reciprocals add to give the reciprocalPower adds PT = P1 + P2 + P3 + ……
4 Solving Problems Set up chart Fill in given information. Fill in what is constant in that type of circuitLook for any place where you either have at least 2 in a column or all but one in a row.Use Ohm’s law to complete columns and characteristics of that type of circuit to complete row.
5 Sample ProblemFour resistors, 3 , 5 , 7 and 9, are arranged in series across a 48-V battery.(a) Draw the circuit diagram.
6 Cont’dFind (b) the equivalent resistance and (c) the current in the circuit. (d) Calculate the voltage drop across each resistor. (e) What power is dissipated in each resistor?You have at least 2 in a column, so V=IRYou have all but one in the resistance row, since it is a series circuit, you add to get the total resistanceSince it is a series circuit, the current is constantYou now have at least 2 in all remaining columns, so use V=IRCalculate all the individual powers using P=VIRecord Given ValuesDouble check with the total power…. Power adds to give the total and P = VI10VT =V1 =V2 =V3 =V4 =IT =I1 =I2 =I3 =I4 =RT =R1 =R2 =R3 =R4 =PT =P1 =P2 =P3 =P4 =4861418222222435799612202836
7 Sample ProblemConnect the same circuit up with the resistors in parallel. (a) Draw the circuit diagram.
8 Cont’d(b) What is the voltage drop across each resistor? (c) What is the equivalent resistance? (d) What current flows through each resistor? (e) Calculate the power dissipated by each resistor.You have all but one in the resistance row, so the reciprocals add to give the reciprocal of the totalSince it is a parallel circuit, voltage is constantNow calculate all of powers using P = VI…… Don’t forget to double check since the individual powers add to give the totalYou have at least 2 in each column so V = IRShow all given information48VT =V1 =V2 =V3 =V4 =IT =I1 =I2 =I3 =I4 =RT =R1 =R2 =R3 =R4 =PT =P1 =P2 =P3 =P4 =4848484837.7184.108.40.206.2735791814768461329256
9 Internal Resistance of a Battery All batteries have internal resistance because some of the energy must be used to drive the current through the battery itself.EMF (electromotive force) also known as the open circuit reading is the maximum amount of energy a battery could produceThe VT (terminal voltage) also known as the closed circuit reading is always less than the EMF due to the internal resistance. It is the actual amount used up in the external circuit.
10 VT = EMF - IriIRE = EMF - IriVT = Terminal Voltage in VoltsEMF = Electromotive force in VoltsI = Current in Ampsri = Internal resistance of the battery in OhmsRE = total resistance of external circuit
11 Cells in SeriesPositive terminal connected to negative terminal of the next batteryEMF addsCurrent is constantInternal resistance addsGives you high voltage for short periods of time
12 Cells in ParallelPositive terminal connected to the positive terminal of the next batteryEMF is constantCurrent addsReciprocals of internal resistances add to give the reciprocal of the totalProvides energy for a long time
13 Sample ProblemThree dry cells each have an EMF of 1.5 V and an internal resistance of 0.1 . What is the EMF if these cells are connected in series?4.5 VWhat is the internal resistance of the battery?0.3 OhmWhat is the line current if this battery is connected to a 10 Ohm resistor?0.437 AmpsWhat is the terminal voltage of the battery?4.37 V
14 Sample ProblemA battery gives an open-circuit reading (EMF) of 3.00 V. The voltmeter is disconnected and the battery is then connected in series with an ammeter and an external load of 11.5 . The ammeter reading is A. Calculate the internal resistance of the battery.0.5 Ohms
15 Complex CircuitsAlso known as combination circuits or networks because parts of the circuit are connected in series and parts in parallelCreate chart… keep simplifying until you have a simple circuit.Solve using rules for series and parallel
16 Sample ProblemFor each circuit below, determine the readings on each meter.R1 = ΩR2 = ΩR3= 6 ΩR4 = 4 ΩV1 = V
17 VT= V1= V2= V3= V4= IT= I1= I2= I3= I4= RT= R1= R2= R3= R4= You can now play the voltage game. Pick a path that has only one unknown on it such as going through R1 and R2 then back to the battery. The total voltage drop on R1 and R2 must be equal to the terminal voltage of the battery. So the voltage drop across R2 is 2 VYou have at least 2 in a column so use V=IRThe current leaving the battery has no choice but to go through R1. So the current through that resistor is the same as ITYou have at least 2 in a column so use V=IRSince R3 and R4 are in parallel with R2, the voltage drop across each one must be the sameRecord the given informationVT=V1=V2=V3=V4=IT=I1=I2=I3=I4=RT=R1=R2=R3=R4=1210222110.170.330.512101264R1 combines with the equivalent resistance of R2, R3, R4 for a combined resistance of 12 Ohms since that is in seriesR2, R3, R4 combine for an equivalent resistance of 2 Ohms since they are in parallel122
18 Kirchhoff’s Laws1st Law - Total current into a junction is equal to the total current leaving the junction…. Also known as the law of conservation of charge
23 Kirchhoff’s Laws2nd Law - The algebraic sum of the changes in potential energy occurring in any closed loop is zero due to the law of conservation of energy.
24 Sign Conventions for 2nd Law Crossing a resistor with current then -IRCrossing a resistor against current then +IRCrossing a battery with current +VCrossing a battery against current -V
25 Sample Problem A Left Clockwise loop starting at A 20I1 +5I2 +30I1 -30 30 Ω30 V20 Ω60 Ω50 V5 ΩI1I2I3ALeft Clockwise loop starting at A20I1+5I2+30I1-30=0You next cross a battery against current so -30Set it equal to zero, you are back at your starting position. The total change in potential around any closed loop is zero.You next cross the 5 Ohm resistor against current so +5I2You next cross the 30 Ohm resistor against current so +30I1You cross the 20 Ohm resistor against current, so +20I1
26 Solving for Currents using K. Laws You must have one first law equation and 2 second law equations.Rearrange equation into proper format.D(I1) + E(I2) + F(I3) = GSet up matrix A (3 x 3) and B (3x1)A-1B
28 50I1 + 5I2 + 0I3 = 300I1 – 5I2 -60I3 = -501I1 – 1I2 + 1I3 = 0Rearrange all equations to fit the DI1 + EI2 + FI3 = G formatCreate a 3 x 3 matrix named A using the left side of each equationMatrix ACreate a 3 x 1 matrix named B using the right side of each equationMatrix B30-50
29 Using the matrices A and B, perform the function A-1B Using the matrices A and B, perform the function A-1B. The output on your calculator screen should beThe top number is the value of I1, the second line is the value of I2, and the last line is the value of I3