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 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 Find (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=IR You have all but one in the resistance row, since it is a series circuit, you add to get the total resistance Since it is a series circuit, the current is constant You now have at least 2 in all remaining columns, so use V=IR Calculate all the individual powers using P=VI Record Given Values Double check with the total power…. Power adds to give the total and P = VI 10 VT = V1 = V2 = V3 = V4 = IT = I1 = I2 = I3 = I4 = RT = R1 = R2 = R3 = R4 = PT = P1 = P2 = P3 = P4 = 48 6 14 18 2 2 2 2 2 24 3 5 7 9 96 12 20 28 36

Sample Problem Connect the same circuit up with the resistors in parallel. (a) Draw the circuit diagram.

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 total Since it is a parallel circuit, voltage is constant Now calculate all of powers using P = VI…… Don’t forget to double check since the individual powers add to give the total You have at least 2 in each column so V = IR Show all given information 48 VT = V1 = V2 = V3 = V4 = IT = I1 = I2 = I3 = I4 = RT = R1 = R2 = R3 = R4 = PT = P1 = P2 = P3 = P4 = 48 48 48 48 37.7 16 9.6 6.9 5.3 1.27 3 5 7 9 1814 768 461 329 256

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. R1 = 10 Ω R2 = 12 Ω R3= 6 Ω R4 = 4 Ω V1 = 12 V

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 V You have at least 2 in a column so use V=IR The current leaving the battery has no choice but to go through R1. So the current through that resistor is the same as IT You have at least 2 in a column so use V=IR Since R3 and R4 are in parallel with R2, the voltage drop across each one must be the same Record the given information VT= V1= V2= V3= V4= IT= I1= I2= I3= I4= RT= R1= R2= R3= R4= 12 10 2 2 2 1 1 0.17 0.33 0.5 12 10 12 6 4 R1 combines with the equivalent resistance of R2, R3, R4 for a combined resistance of 12 Ohms since that is in series R2, R3, R4 combine for an equivalent resistance of 2 Ohms since they are in parallel 12 2

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 I3 = 4 A 1 k I2 = 3 A I1 = 2 A I4 = ? 1A

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 A Left Clockwise loop starting at A 20I1 +5I2 +30I1 -30 30 Ω 30 V 20 Ω 60 Ω 50 V 5 Ω I1 I2 I3 A Left Clockwise loop starting at A 20I1 +5I2 +30I1 -30 =0 You next cross a battery against current so -30 Set 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 +5I2 You next cross the 30 Ohm resistor against current so +30I1 You cross the 20 Ohm resistor against current, so +20I1

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 Bottom junction equation I1 + I3 = I2 30 Ω 30 V 20 Ω 60 Ω 50 V 5 Ω I1 I2 I3 Bottom junction equation I1 + I3 = I2 Left Clockwise Loop Equation: 20I1 +5I2 + 30I1 – 30 =0 Right Clockwise Loop Equation: 50 - 5I2 -60I3 = 0

50I1 + 5I2 + 0I3 = 30 0I1 – 5I2 -60I3 = -50 1I1 – 1I2 + 1I3 = 0 Rearrange all equations to fit the DI1 + EI2 + FI3 = G format Create a 3 x 3 matrix named A using the left side of each equation Matrix A 5 0 0 -5 -60 1 -1 1 Create a 3 x 1 matrix named B using the right side of each equation Matrix B 30 -50

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 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