SPH4UW Kirchhoff’s Laws Lecture timing is fine. No need to rush 1

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

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

Kirchhoff’s Laws e1- I1R1- I2R2-e2=0 (1) Label all currents Choose any direction Choose loop and direction Must start on wire, not element. R4 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. Have students label I5, since it isn’t shown in their drawing e1- I1R1- I2R2-e2=0 For inner loop

Practice ε1- IR1 - ε2 - IR2 = 0 Find I: Label currents Choose loop Write KVR Practice B R1=5 W I Find I: ε1- IR1 - ε2 - IR2 = 0 50 - 5 I - 10 -15 I = 0 I = +2 Amps e1= 50V A R2=15 W e2= 10V What if only went from A to B?, Find VB-VA e2= 10V R1=5 W I e1= 50V R2=15 W A B VB - VA = e1 - IR1 = 50 - 25 = 40 Volts or VB - VA = +IR2 + e2 = 215 + 10 = +40 Volts Therefore B is 40V higher than A

Understanding Resistors R1 and R2 are: R1=10 W Resistors R1 and R2 are: 1) in parallel 2) in series 3) neither E2 = 5 V I2 R2=10 W Definition of parallel: Two elements are in parallel if (and only if) you can make a loop that contains only those two elements. IB - + E1 = 10 V Note that nothing is in series or in parallel! Definition of series: Two elements are in series if (and only if) every loop that Contains R1 also contains R2

Understanding: Voltage Law Practice slide 7 Calculate the current through resistor 1. I1 R1=10 W I1 = 0.5 A I1 = 1.0 A I1 = 1.5 A E2 = 5 V I2 R2=10 W IB <--Start E1 = 10 V Understanding: Voltage Law How would I1 change if the switch was closed? 1) Increase 2) No change 3) Decrease

Starting at Star and move clockwise around loop Understanding slide 7 Calculate the current through resistor 2. I1 R1=10 W I2 = 0.5 A I2 = 1.0 A I2 = 1.5 A E2 = 5 V I2 R2=10 W IB E1 = 10 V Starting at Star and move clockwise around loop

Kirchhoff’s Junction Rule slide 7 Kirchhoff’s Junction Rule Current Entering = Current Leaving I1 = I2 + I3 I1 I2 I3 Understanding I1=1.0A R=10 W IB = I1 + I2 = 1.0A + 0.5 A = 1.5 A IB = 0.5 A IB = 1.0 A IB = 1.5 A E = 5 V R=10 W I2=0.5 IB - + E1 = 10 V

Kirchhoff’s Laws (1) Label all currents Choose loop and direction Choose any direction Choose loop and direction Your choice! R4 R1 E1 R2 R3 E2 E3 I1 I3 I2 I4 R5 A B Write down voltage drops Follow any loops Have them go back to this slide and fill in (5) Write down node equation Iin = Iout

Ohm’s Law

Resistor Series Parallel Wiring Voltage Current Resistance R1 R1 R2 R2 Each resistor on the same wire. Each resistor on a different wire. Wiring Different for each resistor. Vtotal = V1 + V2 Same for each resistor. Vtotal = V1 = V2 Voltage Same for each resistor Itotal = I1 = I2 Different for each resistor Itotal = I1 + I2 Current Increases Req = R1 + R2 Decreases 1/Req = 1/R1 + 1/R2 Resistance

Understanding 1 2 3 R 2R R/2 Which configuration has the smallest resistance? 1 2 3 Which configuration has the largest resistance? 1 2 3

Parallel + Series Tests Resistors R1 and R2 are in series if and only if every loop that contains R1 also contains R2 Resistors R1 and R2 are in parallel if and only if you can make a loop that has ONLY R1 and a loop that ONLY contains R2

Understanding Determine the voltage and current in each resistor First we notice the voltage drop through these two resistor groupings total 10V. Let’s combine to find REQ R3=17W e0=10V R4=5W R2=25W R1=15W Let’s now add R4 to this REQ

Understanding Let’s determine the current, IT e0=10V e0=10V R4: R1: R3=17W e0=10V R4=5W R2=25W R1=15W e0=10V RT=11.04W R1, R2, R3, experience the same voltage This current will flow through each of the resistor groupings R4: R1: R2: R3: Since voltage drop has to total 10V, VEQ=10V-4.5V=5.5V

Determine the Voltage, Current, and Resistance The V-I-R Chart Determine the Voltage, Current, and Resistance 5Ω 7Ω 12V 10Ω Step 1: Fill out the table with known resistors and the Total Voltage for circuit V I R R1 5 R2 7 R3 10 Total 12 V I R R1 R2 R3 Total

Determine the Voltage, Current, and Resistance The V-I-R Chart Determine the Voltage, Current, and Resistance 5Ω 7Ω 12V 10Ω Step 2: Using resistor laws, determine total resistance of circuit. V I R R1 5 R2 7 R3 10 Total 12 9.117 V I R R1 5 R2 7 R3 10 Total 12

Determine the Voltage, Current, and Resistance The V-I-R Chart Determine the Voltage, Current, and Resistance 5Ω 7Ω 12V 10Ω Step 4: Since initial current amount will also pass through resistor R1, we can determine its voltage drop. Step 3: Using Ohm’s law, determine the Current of circuit. V I R R1 6.6 1.32 5 R2 7 R3 10 Total 12 9.117 V I R R1 5 R2 7 R3 10 Total 12 9.117 V I R R1 5 R2 7 R3 10 Total 12 1.32 9.117

Determine the Voltage, Current, and Resistance The V-I-R Chart Determine the Voltage, Current, and Resistance 5Ω 7Ω 12V 10Ω Step 5: Since R2 and R3 have the same Voltage drop, we then must have 12V-6.6V=5.4V. V I R R1 6.6 1.32 5 R2 5.4 7 R3 10 Total 12 9.117 V I R R1 6.6 1.32 5 R2 7 R3 10 Total 12 9.117

Determine the Voltage, Current, and Resistance The V-I-R Chart Determine the Voltage, Current, and Resistance 5Ω 7Ω 12V 10Ω V I R R1 6.6 1.32 5 R2 5.4 7 R3 10 Total 12 9.117 V I R R1 6.6 1.32 5 R2 5.4 0.77 7 R3 0.54 10 Total 12 9.117 Step 6: Use ohm’s law to find currents

Question If a 60 watt light bulb operates at a voltage of 120V, what is the resistance of the bulb? 2Ω 30Ω 240Ω 720Ω 7200Ω

Question Three resistors are connected to a 10-V battery as shown below. What is the current through the 2.0 Ω resistor? 4.0Ω 2.0Ω ε=10V 0.25A 0.50A 1.0A 2.0A 4.0A Since all resistors are in series, the amount of current that passes through any one of them is the same. So we need to simply the circuit to determine that current.

Question Determine the equivalent resistance between points a and b? 0.167Ω 0.25 Ω 0.333 Ω 1.5 Ω 2 Ω 12Ω 3Ω 4Ω 3Ω

Question What is the voltage drop across the 12 ohm resistor in the portion of the circuit shown? 24V 36V 48V 72V 144V 8Ω Since the top branch has 4Ω of resistance and the bottom branch has 12Ω of resistance, therefore 3 times as much current will flow through the 4Ω resistor than the 12Ω resistor. This breaks down the 12A into 9A up and 3A down. So form V=IR we have V=(3A)(12Ω)=36V 4Ω 2Ω 8Ω 12A 12Ω We could also have produced a RT and found 3Ω value, therefore a voltage drop of (3Ω)(12A)=36V on both the upper and lower branch

Question A 100Ω, 120Ω, and 150Ω resistor are connected to a 9-V battery as in the circuit shown below. Which of the three resistors dissipates the most power? We can also answer this question faster by noticing that the voltage drops across the 100Ω and then across the parallel resistors. Since the 100Ω has a higher value than the parallel combination, it will have a larger voltage drop than the combination so via P=IV, it dissipates the most power. The 100Ω resistor The 120Ω resistor The 150Ω resistor Both the 120Ω and the 150Ω All dissipate the same power 100Ω 120Ω 9V 150Ω Therefore the 100Ω resistor dissipates the most amount of power.

Question Given the following circuit 10Ω a 10Ω 20Ω 40Ω ε=120V 100Ω b At what rate does the battery deliver energy to the circuit? Determine the current through the 20 Ω resistor. i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher? Determine the energy dissipated by the 100 Ω resistor in 10 s

Question Given the following circuit At what rate does the battery deliver energy to the circuit? 10Ω 20Ω 100Ω 40Ω ε=120V Recall: Rate is Power RT We need to write this as a simple circuit with a RT so that we can determine the current, I by using V=IR.

Question Given the following circuit b) Determine the current through the 20 Ω resistor. We can determine the voltage drop . Recall: from Question a) we have a 2A current 10Ω 2A 10Ω 120V-(2A)(10Ω)-(2A)(10Ω)-(2A)(10Ω) =60V 20Ω 40Ω The ratio of the resistance of left side to right side is 40:120 or 1:3 ε=120V 100Ω Then we use V=IR to find each individual current. 2A 10Ω Therefore ¼ of 2 A goes through the right and ¾ of 2A goes through the left. Thus (1/4)(2A)=0.5A passes through the 20Ω resistor. OR

Question Given the following circuit c) i) Determine the potential difference between points a and b ii) At which of these two points is the potential higher? 10Ω 20Ω 100Ω 40Ω ε=120V a b i) ii) Point a is at a higher potential. Since current flows from a high potential to a low potential.

Question Given the following circuit d) Determine the energy dissipated by the 100 Ω resistor in 10 s 10Ω 20Ω 100Ω 40Ω ε=120V a b Recall: Energy equals Power multiplied by Time

Understanding Let’s follow a conventional current path through R1 R3=17W e0=10V R4=5W R2=25W R1=15W You can pick any path through the circuit and the total voltage increases and decrease will balance You can reverse the direction of the current and thus the signs, (batteries increase the voltage, resistors drop the voltage) and obtain the same results.

Let us calculate the Current and the Power (used/generated) by the elements of the following circuit. What happens to the Power delivery and consumption if another identical bulb is place in parallel or in series with the first?

Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in parallel. 2A 2A 4A Because the bulbs (resistors) are in parallel, we use the parallel law to determine total resistance of the circuit.

Let us calculate the Current and the Power (used/generated) by the elements of the following circuit when bulbs are in series. 1A Because the bulbs (resistors) are in series, we use the series law to determine total resistance of the circuit.

Question 200Ω ε= 9V 300Ω 400Ω 500Ω Simplify the above circuit so that it consists of one equivalent resistor and the battery. What is the total current through this circuit? Find the voltage across each resistor. Find the current through each resistor. The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same.

Question 200Ω ε= 9V 300Ω 400Ω 500Ω REQ Simplify the above circuit so that it consists of one equivalent resistor and the battery.

Question 200Ω ε= 9V 300Ω 400Ω 500Ω REQ =342.2Ω b) What is the total current through this circuit?

Question 200Ω ε= 9V 300Ω 400Ω 500Ω Find the voltage across each resistor. Find the current through each resistor. Let’s use a VIR chart V I R 200Ω 300Ω 400Ω 500Ω Total 9V 0.0263A 342.2Ω 3.16V 0.0158A 3.16V 0.0105A 5.84V 0.0146A 5.84V 0.0117A

Question 200Ω e) The 500Ω resistor is now removed from the circuit. State whether the current through the 200Ω resistor would increase, decrease, or remain the same. 300Ω ε= 9V 400Ω 500Ω By removing a resistor from a parallel set, we actually increase the resistance of the total circuit. Therefore by Ohm’s law if the voltage remains the same and the resistance increases, the total current must decrease Now through the 200Ω set, the total resistance remains the same, yet the current decreases, therefore the voltage across each resistor decreases as well as the current .