Kirchhoff’s Laws SPH4UW

Last Time Resistors in series: Resistors in parallel: Current thru is same; Voltage drop across is IR i Voltage drop across is same; Current thru is V/R i Last Lecture Solved Circuits What about this one? Today

Kirchhoff’s Rules Kirchhoff’s Voltage Rule (KVR):KVR Sum of voltage drops around a loop is zero. Sum of voltage drops around a loop is zero. Kirchhoff’s Current Rule (KCR): Current going in equals current coming out. Current going in equals current coming out.

Kirchhoff’s Rules I a b=-IR I a b =IR =+E a b a b =-E Between a and b

Kirchhoff’s Laws (1)Label all currents Choose any direction (2)Choose loop and direction Must start on wire, not element. (3)Write down voltage drops -Batteries increase or decrease according to which end you encounter first. -Resistors drop if going with current. -Resistors increase if gong against current. R4R4 I1I1 I3I3 I2I2 I4I4 R1R1 E1E1 R2R2 R3R3 E2E2 E3E3 R5R5 A B  1 - I 1 R 1 - I 2 R 2 -  2 =0 For inner loop

Practice R 1 =5  I ε 1 - IR 1 - ε 2 - IR 2 = I I = 0 I = +2 Amps  1 = 50V R 2 =15   2 = 10V A B What if only went from A to B?, Find V B -V A V B - V A = +IR 2 +  2 = 2  = +40 Volts V B - V A =  1 - IR 1 =  5 = 40 Volts Find I :  2 = 10V R 1 =5  I  1 = 50V R 2 =15  A B Label currents Choose loop Write KVR or Therefore B is 40V higher than A

Understanding R 1 =10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R 2 =10  I2I2 Resistors R 1 and R 2 are: 1) in parallel 2) in series 3) neither + - Definition of parallel: Two elements are in parallel if (and only if) you can make a loop that contains only those two elements. Definition of series: Two elements are in series if (and only if) every loop that Contains R 1 also contains R 2

Practice R 1 =10  E 1 = 10 V IBIB I1I1 R 2 =10  I2I2 1)I 1 = 0.5 A 2)I 1 = 1.0 A 3)I 1 = 1.5 A How would I 1 change if the switch was closed? E 2 = 5 V 1) Increase 2) No change 3) Decrease Calculate the current through resistor 1. Understanding: Voltage Law <-- Start slide 7

Understanding R 1 =10  E 1 = 10 V IBIB I1I1 E 2 = 5 V R 2 =10  I2I2 1)I 2 = 0.5 A 2)I 2 = 1.0 A 3)I 2 = 1.5 A Calculate the current through resistor 2. Starting at Star and move clockwise around loop slide 7

Kirchhoff’s Junction Rule Current Entering = Current Leaving I1I1 I2I2 I3I3 I 1 = I 2 + I 3 1)I B = 0.5 A 2)I B = 1.0 A 3)I B = 1.5 A R=10  E 1 = 10 V IBIB I 1 =1.0A E = 5 V R=10  I 2 = Understanding I B = I 1 + I 2 = 1.0A A = 1.5 A I B = I 1 + I 2 = 1.0A A = 1.5 A slide 7

Kirchhoff’s Laws (1)Label all currents Choose any direction (2)Choose loop and direction Your choice! (3)Write down voltage drops Follow any loops (4)Write down node equation I in = I out R4R4 R1R1 E1E1 R2R2 R3R3 E2E2 E3E3 I1I1 I3I3 I2I2 I4I4 R5R5 A B

You try it! R1R1 R2R2 R3R3 I1I1 I3I3 I2I2 Loop 1: 1.Label all currents 2. Choose loop and direction 3.Write down voltage drops Loop 2: 11 4. Write down node equation Node: 22 In the circuit below you are given ε 1, ε 2, R 1, R 2 and R 3. Find I 1, I 2 and I 3. (Choose any direction) (Current goes +  - for resistor) (Your choice! Include all circuit elements!) Loop 1 Loop 2 +  1 - I 1 R 1 + I 2 R 2 = 0 - I 2 R 2 - I 3 R 3 -  2 = 0 I 1 + I 2 = I 3 3 Equations, 3 unknowns the rest is math!

Calculations R1R1 R2R2 R3R3 I1I1 I3I3 I2I2 11 Loop 1 Loop 2 Loop 1: Loop 2: +  1 - I 1 R 1 + I 2 R 2 = 0 Node: - I 2 R 2 - I 3 R 3 -  2 = 0 I 1 + I 2 = I 3 The negatives only indicate that our current direction choice was wrong.

Practice Circuits R 1 =25 R 2 =100 R 3 =50 I1I1 I3I3 I2I2  1 =12V In the circuit below you are given ε 1, R 1, R 2 and R 3. a)Determine the total resistance of the circuit b)Find I 1, I 2 and I 3. Since R 2 and R 3 are in parallel Now R P and R 1 are in serial This circuit can be broken down into a simple circuit, no need for Kirchhoff Now: This is the current of I 1 The potential, V across R 2 and R 3 is Therefore:

Practice 1.Label all currents In the circuit below, find ε 1, I 2, I 3 (Directions are given) 4Ω4Ω 6Ω6Ω I 1 =0.5 I3I3 I2I2 11 Loop 1 Loop 2 12V 4V 2Ω2Ω 2. Choose loop and direction (Your choice!)

Practice In the circuit below, find ε 1, I 2, I 3 4Ω4Ω 6Ω6Ω I 1 =0.5 I3I3 I2I2 11 Loop 1 Loop 2 12V 4V 2Ω2Ω 3.Write down voltage drops Loop 1: + (0.5A)(2Ω) + ε V- I 2 (4Ω) = 0 Loop 2: + I 2 (4Ω) + 12V-4V + I 3 (6 Ω )= 0 5. Write down node equation Node: 0.5A + I 2 = I 3

Practice In the circuit below, find ε 1, I 2, I 3 4Ω4Ω 6Ω6Ω I 1 =0.5 I3I3 I2I2 Loop 1 Loop 2 12V 4V 2Ω2Ω 1) + (0.5A)(2Ω) + ε V - I 2 (4Ω) = 0 2) + I 2 (4Ω) + 12V - 4V + I 3 (6 Ω )= 0 3) 0.5A + I 2 = I 3 3 Equations, 3 Unknowns 11 The “-” on the currents indicate that our original direction guess was wrong

Practice In the circuit below, find the current in each resistor and the equivalent resistance of the network of five resistors. 2Ω2Ω 13V 1Ω1Ω c 1Ω1Ω 1Ω1Ω 1Ω1Ω I 2 I5I5 I4I4 I 3 I 1 a b d

Practice This “bridge” network cannot be represented in terms of series and parallel combinations. There are five different currents to determine, but by applying the junction rule to junctions a and b, we can determine then in terms of three unknown currents. 2Ω2Ω I 1+ I 2 Loop 1 Loop 3 13V 1Ω1Ω c Loop 2 1Ω1Ω 1Ω1Ω 1Ω1Ω I 2 I 2 + I 3 I 1 – I 3 I 3 I 1 a b d I5I5 I4I4 Using the current directions as guides, we will define 3 loops (3 equations for the 3 unknowns)

Practice 2Ω2Ω I 1+ I 2 Loop 1 Loop V 1Ω1Ω c Loop 2 1Ω1Ω 1Ω1Ω 1Ω1Ω I 2 I 2 + I 3 I 1 – I 3 I 3 I 1 a b d I5I5 I4I4 Loop 1: Loop 2: Loop 3: This is a set of 3 equations and three unknowns. So let’s solve

Practice Loop 1: Loop 2: Loop 3: From loop 3: Substitute this into loop 1 and loop 2 (to eliminate I 2 ) Loop 1: Loop 2: Multiply loop 1 by 5 and adding to loop 2 and solving for I 1 thus and

Practice The total current is: The potential drop across this is equal to the battery emf, namely 13V. Therefore the equivalent resistance of the network is: 2Ω2Ω I 1+ I 2 Loop 1 Loop V 1Ω1Ω c Loop 2 1Ω1Ω 1Ω1Ω 1Ω1Ω I 2 I 2 + I 3 I 1 – I 3 I 3 I 1 a b d I5I5 I4I4