Simple Circuits & Kirchoff’s Rules

Simple Series Circuits  Each device occurs sequentially.  The light dilemma: If light goes all of them go.

Simple Series Circuit - Conservation of Energy  In a series circuit, the of the is equal to. V source + Where we consider the source voltage to be and the voltage drops of each device to be. V source = Since V = ( ): V source =

R1R1 R2R2 + V R3R3 Simple Series Circuit - Conservation of Charge  In a series circuit, the same amount of passes through each device. I T =

Simple Series Circuit – Determining R equivalent  What it the total in a series circuit? Start with of V source = Due to conservation of charge, I Total = I 1 = I 2 = I 3, we can factor out I such that V source = Since V source = : R Total = R Eq =

Simple Parallel Circuit  A parallel circuit exists where components are connected across the same.  Parallel circuits are similar to those used in. + V

Simple Parallel Circuits  Since each device is connected across the : V source = + V

 In parallel circuits, the is equal to the of the through each individual leg. Consider your home plumbing:  Your water comes into the house under pressure.  Each faucet is like a that occupies a leg in the circuit. You turn the valve and the water flows.  The drain reconnects all the faucets before they go out to the septic tank or town sewer.  All the water that flows through each of the faucets adds up to the total volume of water coming into the house as well as that going down the drain and into the sewer.  This analogy is similar to current flow through a parallel circuit. Simple Parallel Circuits Analogy How Plumbing relates to current

Simple Parallel Circuits – Conservation of Charge & Current  The total from the voltage source (pressurized water supply) is equal to the sum of the (flow of water through faucet and drain) in each of the (faucets) I Total = + V

Simple Parallel Circuit – Determining R equivalent  What it the total resistance in a parallel circuit? Using conservation of charge I Total = or Since V source = V = V = V we can substitute V source in (1) as follows

Simple Parallel Circuit – Determining R equivalent  What it the total resistance in a parallel circuit (cont.)? However, since I Total = / substitute in (2) as follows Since V source cancels, the relationship reduces to Note: R total has been replaced by.

Kirchoff’s Rules  Loop Rule (Conservation of ): The sum of the ( )equals the sum of the ( ) around a closed loop.  Junction Rule (Conservation of Electric ): The sum of the magnitudes of the going into a junction equals the sum of the magnitudes of the leaving a junction.

Rule #1: Voltage Rule (Conservation of ) R1R1 R2R2 + V R3R3 ΣVΣV V source – V 1 – V 2 – V 3 =

Rule #2: Current Rule (Conservation of Electric ) I1I1 I3I3 I2I2 I 1 + I 2 + I 3 =

Example Using Kirchoff’s Laws  Create individual loops to analyze by Kirchoff’s.  Arbitrarily choose a direction for the current to flow in each loop and apply Kirchoff’s. + + R 3 = 5Ω  2 = 5V I1I1 I2I2 I3I3  1 = 3V R 1 = 5Ω R 2 = 10Ω

Ex. (cont.)  Apply Kirchoff’s Current Rule ( = ): I 1 + I 2 = (1)  Apply Kirchoff’s Voltage Rule to the left loop (Σv = ):  1 – – = Substitute (1) to obtain:  1 – – ( ) = (2)

Ex. (cont.)  Apply Kirchoff’s Voltage Rule to the right loop:  2 – – = Substitute (1) to obtain:  2 – – ( ) = (3)

Ex. (cont.)  List formulas to analyze. I 1 + I 2 = (1)  1 – – ( ) = (2)  2 – – ( ) = (3)  Solve 2 for I 1 and substitute into (3)  1 – – – = – – = –  1 I 1 ( ) =  1 -  1 - ( ) I 1 =

( ) ( )  Plug in known values for R 1, R 2, R 3,  1 and  2 and then solve for I 2 and then I 3. Ex. (cont.)  2 – – + =  2 ( ) – ( ) – + – ( ) = 0 [ [ ( ) – =  2 – – [ [ Multiply by (R 1 + R 2 ) to remove from denominator. I 2 = A

Ex. (cont.)  Plug your answer for I 2 into either formula to find I 1  1 – – ( ) =  What does the tell you about the current in loop 1? I 1 =  1 - ( ) I 1 = 3V – ( )( ) ( + ) I 1 =

Ex. (cont.)  Use formula (1) to solve for I 3 I 1 + I 2 = + =

How to use Kirchhoff’s Laws A two loop example: Analyze the circuit and identify all circuit nodes and use KCL. (2)  1  I 1 R 1  I 2 R 2 = 0 (3)  1  I 1 R 1   2  I 3 R 3 = 0 (4) I 2 R 2   2  I 3 R 3 = 0  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 (1) I 1 = I 2 + I 3 Identify all independent loops and use KVL.

How to use Kirchoff’s Laws  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3 Solve the equations for I 1, I 2, and I 3 : First find I 2 and I 3 in terms of I 1 :  From eqn. (2) From eqn. (3) Now solve for I 1 using eqn. (1):

Let’s plug in some numbers  1  2 R1R1 R3R3 R2R2 I1I1 I2I2 I3I3  1 = 24 V  2 = 12 V R 1 = 5  R 2 =3  R 3 =4  Then, and I 1 =2.809 A I 2 = A, I 3 = A