Remember that for series circuits Kirchoffs laws must

Remember that for series circuits… Kirchoff’s laws must hold! 1. The sum of the currents flowing into any point in a circuit is equal to the sum of the currents flowing out of that point. 2. The sum of the e. m. f. s around any loop in a circuit is equal to the sum of the p. d. s around the loop. VT VT = V 1 + V 2 Now consider this same circuit but with the resistors in different positions. (next slide) A V 1 V 2

Internal Resistance (r) Most power supplies have some amount of resistance due to its internal structure (or composition). This means that some of the ‘voltage’ that can be supplied by the cell, will be used up across its own internal resistance. n This can never be harnessed, and is considered to be “lost volts”. n [E – lost volts] = the useful volts; this will be available to be used in the external circuit (across R in this case). n E = 24 V r= 4 W r What is the current that is drawn from the cell? I = V/R = 24/(4+10) R = 10 W So I = 24/14 = 1. 7 A

Internal Resistance (r) Calculate the lost volts and the useful voltage. Lost volts = Ir = 1. 7 x 4 = 6. 8 V Useful p. d. = IR = 1. 7 x 10 = 17 V Generally E = 24 V r= 4 W r R = 10 W Current drawn = 1. 7 A E = IR + Ir , E = the emf of the cell which is constant. IR = the useful p. d. or the terminal p. d with a closed circuit (ie the reading of the voltmeter across the terminals of the power supply) = E - Ir

Under what conditions will E = terminal p. d. ? If E = Ir + IR, and E – Ir = terminal p. d, then when I = 0 A, E = the terminal p. d. E=8 V This means that if you would like to know the emf of a cell, you should attach a voltmeter across it, without attaching an external circuit. With switch S open, the reading on the ammeter = 0 A, so (theoretically) lost volts = 0 V. Practically, only an insignificant current is drawn (since voltmeters have extremely high resistances), and the lost p. d. is negligible. Once the switch is closed, a large current will flow through the entire circuit, Ir will increase, and the reading on the voltmeter will drop. V S A

Analyzing the Circuit…using Ohm’s Law When we attach the resistor to the DC voltage source, current begins to flow . 5 K Ω 1. 5 V n How much current will flow? Ohm’s Law (V=IR) ->Describes the relationship between the voltage (V), current (I), and resistance (R) in a circuit 0 V Using Ohm’s Law, we can determine how much current is flowing through our circuit

Analyzing the Circuit using Ohm’s Law How much current will flow? I = 3 m. A . 5 K Ω 1. 5 V Use Ohm’s Law: V =I x R 1. 5 V = I x. 5 K Ω Solve for I: I = 1. 5 V /. 5 KΩ = 3 m. A n 0 V So, 3 m. A will flow through the. 5 kΩ resistor, when 1. 5 Volts are across it

Resistors in Series R 1 =. 5 K Ω 1. 5 V Req R 2 = 1 K. 5 KΩΩ Resistors connected by only 1 terminal, back-to-back, are considered to be in ‘series’ We can replace the two series resistors with 1 single resistor, we call Req The value of Req is the SUM of R 1 & R 2: Req=R 1+R 2=. 5 K Ω +. 5 K Ω = 1 KΩ 0 V

Resistors in Series I = 1. 5 m. A 1. 5 V Req = 1 K Ω Usewe Ohm’s Law: the n. Now can find V through = I x the Reqcircuit current using 1. 5 VOhm’s = I Law x 1 K Ω Solve for I: I = 1. 5 V / 1 K Ω = 1. 5 m. A 0 V The bigger the resistance in the circuit, the harder it is for current to flow

I = 1. 5 m. A Resistors in Series Back to our original series circuit, with R 1 and R 2 The current is the SAME through each resistor R 1 =. 5 K Ω 1. 5 V R 2 =. 5 K Ω 0 V Current flows like water through the circuit, notice how the 1. 5 m. A ‘stream of current’ flows through both resistors equally Ohm’s Law shows us voltage across each resistor: V(R 1) = 1. 5 m. A x. 5 K Ω =. 75 V V(R 2) = 1. 5 m. A x. 5 K Ω =. 75 V

Resistors in Parallel R 1 =. 5 K Ω 1. 5 V R 2 =. 5 K Ω Resistors connected at 2 terminals, sharing the same node on each side, are considered to be in ‘parallel’ Unlike before, we cannot just add them. We must add their inverses to find Req: 0 V Req =. 25 K Ω

Resistors in Parallel This is the equivalent circuit I = 6 m. A n 1. 5 V Req =. 25 K Ω Use Ohm’s Law, we find the current through Req: V = I x Req 1. 5 V = I x. 25 K Ω Solve for I: I = 1. 5 V /. 25 KΩ = 6 m. A 0 V The smaller the resistance in the circuit, the easier it is for current to flow

Resistors in Parallel Back to our original series circuit, with R 1 and R 2 1. 5 V R 1 =. 5 K Ω R 2 =. 5 KΩ 0 V The current is NOT the SAME through all parts of the circuit Current flows like water through the circuit, notice how the 6 m. A ‘stream of current’ splits to flow into the two resistors The Voltage across each resistor is equal when they are in parallel

Resistors in Parallel The voltage is 1. 5 V across each resistor I = 3 m. A I = 6 m. A 1. 5 V R 1 =. 5 K Ω Ohm’s Law tells us the current through each: I(R 1)=V/R= 1. 5 V /. 5 KΩ = 3 m. A I(R 2)=V/R= 1. 5 V /. 5 KΩ = 3 m. A R 2 =. 5 K Ω 0 V The 6 m. A of current has split down the two legs of our circuit It splits equally between the two legs, because the resistors have the same value The current will split differently if the resistors are not equal…

Resistors in Parallel I = 6 m. A This is the equivalent circuit n 1. 5 V Req =. 25 K Ω Use Ohm’s Law, we find the current through Req: V = I x Req 1. 5 V = I x. 25 K Ω Solve for I: I = 1. 5 V /. 25 K Ω = 6 m. A 0 V The smaller the resistance in the circuit, the easier it is for current to flow

NOTATION: NODE VOLTAGES The voltage drop from node X to a reference node (ground) is called the node voltage Va , Vb Example: a b + Va _ + + _ Vb _ ground

KIRCHHOFF’S VOLTAGE LAW

KVL EXAMPLE Examples of three closed paths: a v 2 1 + va Path 2: Path 3: + vb - 3 Path 1: b v 3 c 2 + vc

KIRCHHOFF’S CURRENT LAW Applying conservation of current. n n The net current entering a node is zero Alternatively, the sum of the currents entering a node equals the sum of the currents leaving a node

KIRCHOFF’S CURRENT LAW USING SURFACES Another example Example surface 50 m. A 5 m. A entering 2 m. A i i=7 m. A leaving i? i must be 50 m. A

GENERALIZATION OF KCL Sum of currents entering/leaving a closed surface is zero Could be a big chunk of circuit in here, e. g. , could be a “Black Box” Note that circuit branches could be inside the surface. The surface can enclose more than one node!

KIRCHHOFF’S CURRENT LAW EXAMPLE Currents entering the node: 24 A Currents leaving the node: 4 A + 10 A + i Three formulations of KCL:

How would the readings on the ammeter and the voltmeter be affected if a second resistor, identical to the first, was added in series. The current will be affected by the total resistance in the circuit. 12 V 2 A A If the resistance is doubled, the electrons will move twice as slowly so the new current reading will be 1 A Kirchoff’s 1 st Law V If the charges are flowing slower less work will be done 12 V as they move through each resistor, the voltage provided by the cell will be shared. The reading on the voltmeter will drop to 6 V. Kirchoff’s 2 nd Law Which of Kirchoff’s laws can be related to these statements?

How would the readings on the ammeter (initially 2 A) and the voltmeter (initially 12 V) be affected if a second resistor, identical to the first was added in parallel? Every electron will have 12 V to 12 V use to do work, but it has a choice of two paths to take. A The voltmeter reading will still V read 12 V. The voltage across the second branch of this parallel circuit will also be 12 V. Kirchoff’s 2 nd The cell in this circuit will run down faster than one attached to a series circuit. It is, essentially, being used by two circuits at the same time. The current drawn will be twice as large. The ammeter will read 4 A. This will split up and 2 A will then flow through each branch. Kirchoff’s 1 st Which of Kirchoff’s laws can be related to these statements?

Using KVL, KCL, and Ohm’s Law to Solve a Circuit