Simple Circuits Kirchoffs Rules Simple Series Circuits o

Simple Circuits & Kirchoff’s Rules

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

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

Simple Series Circuit Conservation of Charge o In a series circuit, the same amount of passes through each device. IT = R 1 V+ R 2 R 3

Simple Series Circuit – Determining Requivalent o What it the total a series circuit? n n n in Start with of Vsource = Due to conservation of charge, ITotal = I 1 = I 2 = I 3, we can factor out I such that Vsource = Since Vsource = : RTotal = REq =

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

Simple Parallel Circuits o Since each device is connected across the : Vsource = V+

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

Simple Parallel Circuits – Conservation of Charge & Current o 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) ITotal = V+

Simple Parallel Circuit – Determining Requivalent o What it the total resistance in a parallel circuit? n n Using conservation of charge ITotal = or Since Vsource = V = V we can substitute Vsource in (1) as follows

Simple Parallel Circuit – Determining Requivalent o What it the total resistance in a parallel circuit (cont. )? n n However, since ITotal = in (2) as follows / substitute Since Vsource cancels, the relationship reduces to Note: Rtotal has been replaced by .

Kirchoff’s Rules o Loop Rule (Conservation of n o ): The sum of the ( )equals the sum of the ( ) around a closed loop. Junction Rule (Conservation of Electric ): n 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 R 1 V+ ΣV R 2 R 3 Vsource – V 1 – V 2 – V 3 = )

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

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

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

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

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

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

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

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

How to use Kirchhoff’s Laws A two loop example: R 1 I 3 I 2 I 1 e 1 R 2 e 2 R 3 • Analyze the circuit and identify all circuit nodes and use KCL. (1) I 1 = I 2 + I 3 • Identify all independent loops and use KVL. (2) e 1 - I 1 R 1 - I 2 R 2 = 0 (3) e 1 - I 1 R 1 - e 2 - I 3 R 3 = 0 (4) I 2 R 2 - e 2 - I 3 R 3 = 0

How to use Kirchoff’s Laws R 1 I 1 e 1 I 3 I 2 e 2 R 3 • 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): Þ

Let’s plug in some numbers R 1 I 1 e 1 = 24 V Then, I 3 I 2 e 2 R 2 e 2 = 12 V R 3 R 1= 5 W R 2=3 W R 3=4 W and I 1=2. 809 A I 2= 3. 319 A, I 3= -0. 511 A