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Series and Parallel Circuits Characteristics of Series Circuits Only one path (electron has no choices, must go through all components) If one goes out, they all go out Voltage adds VT=V1+V2+V3+…. Current is constant IT=I1=I2=I3=…. Resistance adds RT=R1+R2+R3+…. Power adds PT=P1+P2+P3+…. Characteristics of Parallel Circuits More than one path Voltage is constant VT=V1=V2=V3=…. Current adds IT=I1+I2+I3+…. Resistance reciprocals add to give the reciprocal 1 1 1 1 .... RT R1 R2 R3 Power adds PT = P1 + P2 + P3 + …… Solving Problems Set up chart Fill in given information. Fill in what is constant in that type of circuit Look 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. Sample Problem Four resistors, 3 , 5 , 7 and 9, are arranged in series across a 48-V battery. (a) Draw the circuit diagram. Cont’d You have(b) at but Find the equivalent resistance and (c) You have all one in thethe Since it is a series circuit, Record Given Values You Calculate now have all the at individual least 2 in all leastcheck 2 in row, a withsince Double the total Power resistance itthe is apower…. current is constant the current in circuit. (d) Calculate the remaining powers using columns, P=VI so use column, so adds to give the you totaladd andtoP get = VI series circuit, V=IR voltage drop across each resistor. (e) What V=IR the total resistance power is dissipated in each resistor? VT = 48 V1 = 6 V2 = 10 V3 = 14 V4 = 18 IT = 2 I1 = 2 I2 = 2 I3 = 2 I4 = 2 RT = 24 R1 = 3 R2 = 5 R3 = 7 R4 = 9 PT = 96 P1 = 12 P2 = 20 P3 = 28 P4 = 36 Sample Problem Connect the same circuit up with the resistors in parallel. (a) Draw the circuit diagram. Cont’d (b)have isallthe voltage across each You but one in the voltage resistance row, so the You have atparallel least 2 circuit, in each column Since Now calculate itWhat is aall of powers usingdrop Pis=constant VI…… Don’t reciprocals add(c) to give the is reciprocal of thepowers total add to resistor? What the equivalent so V= Show all given information forget toIR double check since the individual give the total resistance? (d) What current flows through each resistor? (e) Calculate the power dissipated by each resistor. VT = 48 V1 = 48 V2 = 48 V3 = 48 V4 = 48 IT = 37.7 I1 = 16 I2 = 9.6 I3 = 6.9 I4 = 5.3 RT = 1.27 R1 = 3 R2 = 5 R3 = 7 R4 = 9 PT =1814 P1 = 768 P2 = 461 P3 = 329 P4 = 256 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 produce The 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. VT = EMF - Iri IRE = EMF - Iri VT = Terminal Voltage in Volts EMF = Electromotive force in Volts I = Current in Amps ri = Internal resistance of the battery in Ohms RE = total resistance of external circuit Cells in Series Positive terminal connected to negative terminal of the next battery EMF adds Current is constant Internal resistance adds Gives you high voltage for short periods of time Cells in Parallel Positive terminal connected to the positive terminal of the next battery EMF is constant Current adds Reciprocals of internal resistances add to give the reciprocal of the total Provides energy for a long time Sample Problem Three 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 V What is the internal resistance of the battery? 0.3 Ohm What is the line current if this battery is connected to a 10 Ohm resistor? 0.437 Amps What is the terminal voltage of the battery? 4.37 V Sample Problem A 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 0.250 A. Calculate the internal resistance of the battery. 0.5 Ohms Complex Circuits Also known as combination circuits or networks because parts of the circuit are connected in series and parts in parallel Create chart… keep simplifying until you have a simple circuit. Solve using rules for series and parallel Sample Problem For each circuit below, determine the readings on each meter. V1 = 12 V R1 = 10 Ω R2 = 12 Ω R3= 6Ω R4 = 4Ω You can now play the voltage game. Pick a path that has only You have at 2 inasa going column unknown onleast it such R2 then back Theone current leaving the battery has nothrough choiceRbut to go 1 and You have at least 2 ininformation column with so useR2V=IR Since R3V=IR and R4 are ina parallel , the Record the given so use to theRbattery. total through voltage drop on R and R must be through theThe current that resistor is the 1. So voltage drop across each one must be the1 same 2 equal the terminal voltage of the battery. So the voltage drop same as Ito T across R2 is 2 V VT= V1= 12 IT= V2= 10 I 1= 1 RT= 1 R1Rcombines with the for 2, R3, R4 combine equivalent an equivalent resistance of R2, R3,resistance R4 for a combined of 2 Ohms resistance since they of 12 areOhms in since parallel that is in series I3= 0.17 R2= 10 2 2 I2= R1= 12 V3= 12 2 I4= 0.33 R3= 0.5 R4= 6 2 12 V4= 4 Kirchhoff’s Laws 1st Law - Total current into a junction is equal to the total current leaving the junction…. Also known as the law of conservation of charge Sample Problem Find I4 I1 = 2 A 1 k I2 = 3 A 1 k I3 = 4 A 1 k I4 = ?1A 1 k Kirchhoff’s Laws 2nd 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. Sign Conventions for 2nd Law Crossing a resistor with current then -IR Crossing a resistor against current then +IR Crossing a battery with current +V Crossing a battery against current -V Sample Problem 20 Ω 60 Ω A 30 V Left Clockwise loop starting at A 50 V I1 I3 I2 5Ω 30 Ω 20I1 +5I2 +30I1 -30 =0 You Set You itnext next equal cross cross to a20 the battery 5you Ohm against areresistor backcurrent atagainst your You next cross the 30 Ohm resistor against You cross thezero, Ohm resistor against so starting current -30 so position. soso+30I +5I 21 The total change in current current, +20I 1 potential around any closed loop is zero. 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) = G Set up matrix A (3 x 3) and B (3x1) A-1B Sample Problem 20 Ω 30 V 60 Ω Bottom junction equation 50 V I1 I3 I2 5Ω I1 + I 3 = I2 30 Ω Left Clockwise Loop Equation: Right Clockwise Loop Equation: 20I1 +5I2 + 30I1 – 30 =0 50 - 5I2 -60I3 = 0 50I1 + 5I2 + 0I3 = 30 Rearrange all equations to fit the DI1 + EI2 + FI3 = G format 0I1 – 5I2 -60I3 = -50 1I1 – 1I2 + 1I3 = 0 Create a 3 x 3 matrix named A using the left side of each equation Create a 3 x 1 matrix named B using the right side of each equation Matrix A 50 5 0 -5 -60 1 -1 Matrix B 30 -50 0 0 1 Using the matrices A and B, perform the function A-1B. The output on your calculator screen should be 0.47887 1.21126 0.73239 The top number is the value of I1, the second line is the value of I2, and the last line is the value of I3