ELECTRICAL CIRCUITS All you need to be an

ELECTRICAL CIRCUITS All you need to be an inventor is a good imagination and a pile of junk. -Thomas Edison

Ohm’s Law I=V/R Georg Simon Ohm (1787 -1854) I = Current (Amperes) (amps) V = Voltage (Volts) R = Resistance (ohms)

How you should be thinking about electric circuits: Voltage: a force that pushes the current through the circuit (in this picture it would be equivalent to gravity)

How you should be thinking about electric circuits: Resistance: friction that impedes flow of current through the circuit (rocks in the river)

How you should be thinking about electric circuits: Current: the actual “substance” that is flowing through the wires of the circuit (electrons!)

Would This Work?

The Central Concept: Closed Circuit

circuit diagram Scientists usually draw electric circuits using symbols; cell lamp switch wires

Simple Circuits n Series circuit n n All in a row 1 path for electricity 1 light goes out and the circuit is broken Parallel circuit n n Many paths for electricity 1 light goes out and the others stay on

1 2

PARALLEL CIRCUIT n n n Place two bulbs in parallel. What do you notice about the brightness of the bulbs? Add a third light bulb in the circuit. What do you notice about the brightness of the bulbs? Remove the middle bulb from the circuit. What happened?

measuring current Electric current is measured in amps (A) using an ammeter connected in series in the circuit. A

measuring current This is how we draw an ammeter in a circuit. A A SERIES CIRCUIT PARALLEL CIRCUIT

measuring voltage The ‘electrical push’ which the cell gives to the current is called the voltage. It is measured in volts (V) on a voltmeter V

measuring voltage This is how we draw a voltmeter in a circuit. V SERIES CIRCUIT V PARALLEL CIRCUIT

OHM’s LAW Measure the current and voltage across each circuit. n Use Ohm’s Law to compute resistance Series Circuit n Voltage Current Resistance Parallel Circuit Voltage Current Resistance

measuring current SERIES CIRCUIT • current is the same at all points in the circuit. 2 A 2 A 2 A PARALLEL CIRCUIT • current is shared between the components 2 A 2 A 1 A 1 A

fill in the missing ammeter readings. 3 A ? 4 A ? 4 A 3 A 1 A ? 1 A 1 A ?

The circuit is no longer complete, therefore current can not flow The voltage decreases because the current is decreased and the resistance increases.

The current remains the same. The total resistance drops in a parallel circuit as more bulbs are added The current increases.

Series and Parallel Circuits n Series Circuits n n n only one end of each component is connected e. g. Christmas tree lights Parallel Circuits n n both ends of a component are connected e. g. household lighting

copy the following circuits and fill in the missing ammeter readings. 3 A ? 4 A ? 4 A 3 A 1 A ? 1 A 1 A ?

measuring voltage Different cells produce different voltages. The bigger the voltage supplied by the cell, the bigger the current. Unlike an ammeter, a voltmeter is connected across the components Scientist usually use the term Potential Difference (pd) when they talk about voltage.

measuring voltage V V

series circuit • voltage is shared between the components 3 V 1. 5 V

parallel circuit • voltage is the same in all parts of the circuit. 3 V 3 V 3 V

measuring current & voltage copy the following circuits on the next two slides. complete the missing current and voltage readings. remember the rules for current and voltage in series and parallel circuits.

measuring current & voltage a) 6 V 4 A V A

measuring current & voltage b) 4 A 6 V A V A

answers a) b) 4 A 6 V 4 A 4 A 3 V 6 V 6 V 2 A 3 V 4 A 4 A 6 V 2 A

Voltage, Current, and Power n n n One Volt is a Joule per Coulomb (J/C) One Amp of current is one Coulomb per second (6. 24 x 10^18 electrons/second). If I have one volt (J/C) and one amp (C/s), then multiplying gives Joules per second (J/s) n n this is power: J/s = Watts So the formula for electrical power is just: P = VI: power = voltage current

Kirchoff’s Laws n n These laws add up to nothing! Yet they completely characterize circuit behavior. Kirchoff’s Voltage Law (KVL) - The sum of voltages taken around any loop is zero. n n The start and end points are identical; consequently there is no potential difference between them. Kirchoff’s Current Law (KCL) – The sum of currents entering any node is zero. n A consequence of the law of conservation of charge.

Circuit components n Active vs. Passive components n n n Lumped vs. Distributed Constants n n n Active ones may generate electrical power. Passive ones may store but not generate power. Distributed constant components account for propagation times through the circuit branches. Lumped constant components ignore these propagation times. Appropriate for circuits small relative to signal wavelengths. Linear, time invariant (LTI) components are those with constant component values.

Active circuit components n n Conservation of energy: active components must get their power from somewhere! From non-electrical sources n n Batteries (chemical) Dynamos (mechanical) Transducers in general (light, sound, etc. ) From other electrical sources n n n Power supplies Power transformers Amplifiers

Passive lumped constants n Classical LTI n Resistors are AC/DC components. Inductors are AC components (DC short n Capacitors are AC components (DC open n n circuit). Other components n n n Rectifier diodes. Three or more terminal devices, e. g. transistors. Transformers.

DC circuits n The basic LTI component is the Resistor n n n Customarily represented by R. The SI unit is the Ohm [� ]. Ohm’s Law: V = I R Ohm’s and Kirchoff’s laws completely prescribe the behavior of any DC circuit comprising LTI components.

AC circuits -- Components n Basic LTI components n n Resistor, R, [� ] (Ohms) Inductor, L, [H] (Henrys) Capacitor, C, [F] (Farads) Frequency n n Repetition rate, f, [Hz] (Hertz) Angular, �= 2� f, [1/s] (radians/sec)

AC Components: Inductors Current in an inductor generates a magnetic field, B = K 1 I n Changes in the field induce an inductive voltage. V = K 2 (d. B/dt) n The instantaneous voltage is V = L(d. I/dt), where L = K 1 K 2. This is the time domain behavior of an inductor. n

AC Components: Capacitors Charge in a capacitor produces an electric field E, and thus a proportional voltage, Q = C V, Where C is the capacitance. n The charge on the capacitor changes according to I = (d. Q/dt). n The instantaneous current is therefore I = C(d. V/dt). This is the time domain behavior of a capacitor. n

AC circuits -- Impedance n Impedance and Ohm’s Law for AC: n n Impedance is Z = R + j. X, where j = � -1, and X is the reactance in [� ]. Ohm’s AC Law in s domain: v = i Z Resistance R dissipates power as heat. Reactance X stores and returns power. n n Inductors have positive reactance Xl=� L Capacitors have negative reactance Xc=1/� C

Impedance shortcuts • The impedance of components connected in series is the complex sum of their impedances. n The impedance of components connected in parallel is the reciprocal of the complex sum of their reciprocal impedances.

METHODS OF CIRCUIT ANALYSIS 45

Methods of Circuit Analysis n n Mesh Analysis Nodal Analysis 46

Mesh Analysis n n Kirchhoff’s Voltage Law (KVL) forms the basis of mesh analysis. This technique is applicable to n n n Basic circuit Circuit with dependent source Circuit with current source n n Case 1: Current source at the outer most boundary (known as mesh current) Case 2: Current source in between two loops (known as supermesh) 47

Step to determine Mesh Current n n n Assign mesh currents I 1, I 2…, In to the n meshes Apply KVL to each of n meshes. Use Ohm’s Law to express voltages in terms of mesh currents. Solve the resulting n simultaneous equation to get the mesh current, 48

Example 10. 3 For the circuit below, find Io using mesh analysis 49

Solution Applying KVL to Mesh 1 …(1) Mesh 2 …(2) Substitute (I 3=5) into meshes (1) and (2) …(3) …(4) 50

Solution Put equation (3) and (4) in matrix form Find determinant for the matrix (Cramer’s Rule) 51

Solution Use Cramer’s rule to solve for I 2 Hence Io = (-I 2) = 52

Practice Problem 10. 3 For the circuit below, find Io using mesh analysis 53

Solution 54

Solution Mesh 1 …(1) Mesh 2 Mesh 3 Insert Mesh 3 into Mesh 2 …(2) 55

Solution Simplify Equation (1) …(3) Substitute equation (3) into (2) 56

Solution Hence 57

Example 10. 4 For the circuit below, find Vo using mesh analysis 58

Solution 59

Solution Mesh 1 …(1) Mesh 2 Supermesh …(2) Due to current source between meshes 3 and 4 at node A …(3) 60

Solution Combine I 2 = -3 into equation (1) …(4) Combine I 2 = -3 into equation (2) and (3) …(5) Put equation (4) and (5) into matrix 61

Solution Use Cramer’s Rule to solve for I 1 62

Solution Solve for Vo 63

Practice Problem 10. 4 64

Solution 65

Solution Mesh 1 …(1) Supermesh …(2) Also the current source between meshes 2 and 3 …(3) 66

Solution Eliminating I 3 from equation (1) and (2) …(4) …(5) Put equation (4) and (5) into matrix 67

Solution Use Cramer’s Rule to solve for I 1 and then Io 68

Exercise III (Problem 10. 38) Using mesh analysis, find Io 69

Solution 70

Solution Mesh 1 …(1) Mesh 2 …(2) Substitute (1) into (2) …(3) 71

Solution Supermesh …(4) …(5) Substitute (1) and (5) into (4) …(6) 72

Solution Put equation (3) and (6) into matrix Use Cramer’s Rule to solve for I 2 73

Nodal Analysis n n The basis of nodal analysis is Kirchhoff’s Current Law (KCL). This technique is applicable to n n n Basic Circuit with dependent source Circuit with voltage source n n Case 1: Voltage source in between reference node and essential node Case 2: voltage source in between two nodes 74

Step to determine Node Voltages n n Select a node as the reference node. Assign voltages V 1, V 2…, Vn-1 to the remaining n-1 nodes. Apply KCL to each of the n-1 nonreference node. Use Ohm’s Law to express the branch currents in term of node voltages. Solve the resulting simultaneous equation to obtain the unknown node voltage. 75