ECE 102 Engineering Computation Chapter 6 Circuit Elements Dr. Herbert G. Mayer, PSU Status 10/11/2015 For use at CCUT Fall 2015

Syllabus Electric Charge and Field Voltage Current Resistor Inductor Capacitor Voltage and Current Sources Voltage and Current Dividers Examples

2 Electric Charge Electric charge ( q or Q ) is an intrinsic property of certain subatomic particles. A particle’s charge affects its motion in the presence of electric and magnetic fields. The SI unit for electric charge is the coulomb (C). 1 C = 1 A · s The amount of charge possessed by a single electron is ≈ 1.602× C. Charge can be positive or negative.

3 Electric Field Lorentz Force Law → The force F acting on a charged particle in the presence of an electric field and a magnetic field is: If the magnetic field is zero, then the electric field that pervades the space around the charge is: SI units for electric field: V/m (1 N/C = 1 V/m) F, E, B, & v are vectors q = electric charge of the particle v = velocity of the particle E = electric field, B = magnetic field FqFq E =

4 Voltage If there is a difference in electric potential between two spatial points, then a non-zero electric field will exist between them. An electric field causes charged particles within the field to move. The voltage V is the amount of work done in moving a charge. SI unit for voltage: volt (V) (1 V = 1 J/C)

5 Current Current is the continuous movement of charge in a particular direction. Electric current I is the rate at which charge flows through a cross-sectional area A. Average: A I dq dt I = ΔqΔtΔqΔt I ave = Instantaneous:

6 Charge carriers may be:  electrons (–)  ions (+)  holes (+) Direct current (DC) → Charge carriers move in one direction only (macroscopic behavior). Alternating current (AC) → The charge carrier direction varies periodically with time. SI unit for current: ampere (A) (1 A = 1 C/s)

7 Conventional Current vs. Electron Current Conventional Current  Positive charge carriers flow from high to low potential (same direction as the electric field). Electron Current  Negative charge carriers flow from low to high potential (opposite direction to the electric field). In circuit analysis, conventional current is assumed, even if electrons are the primary charge carriers. V high V low – + E qV high qV low + – Energy level representation

8 V high V low Circuit + – + – + – + – –– ++ Positive charge moving: From V high to V low → Energy is dissipated From V low to V high → Battery supplies energy Negative charge moving: From V low to V high → Energy is dissipated From V high to V low → Battery supplies energy Battery

Resistance The resistance R is a measure of the opposition to direct current through a material. Interactions of charge carriers with the structure of the material impedes the current. Classes of materials:  Conductor (low R : e.g., many metals)  Insulator (high R : e.g., ceramic)  Semiconductor (intermediate R, e.g. doped Si) SI unit for resistance: ohm ( Ω ) 9

10 Ohm’s Law At a constant temperature, the current I through certain materials is directly proportional to the potential difference ΔV between its ends. The resistance R is defined as: The general form of Ohm’s Law is: I  ΔVI  ΔV

11 Example: Note: It is understood that Ohm’s Law refers to a potential difference. The Δ is usually omitted. R = 2 Ω I V1V1 V2V2 V1V1 V2V2 I 5 V0 V2.5 A 5 V2 V1.5 A 2 V5 V–1.5 A 1 V–3 V2 A 3 V 0 A If the potential difference ΔV is zero, no current flows through the resistor.

12 Application of Ohm’s Law Given:Material of known resistance R Voltage V is applied across the material Result:Current I = V / R will flow through it. Given:Material of known resistance R Known current I flowing through it Result:Voltage V = I · R exists across the material (known as a “voltage drop”). Given:Known voltage V across the material Known current I through the material Result:Resistance of the material is R = V / I.

13 Conductance The conductance G is a measure of the ease with which direct current flows through a material. (reciprocal of resistance) Ohm’s Law in terms of conductances: SI unit for conductance: siemens (S)

14 Power Power is the rate at which energy is generated or dissipated by an electrical element. where V = Voltage (V or J/C) I = Current (A or C/s) R = Resistance (Ω) SI unit for power: watt (W) (1 W = 1 J/s)

15 Example: What is the voltage drop across the resistor? What is the value V 1 ? What is the power dissipated by the resistor? R = 4.0 Ω I = 0.25 A V1V1 V 2 = 6.0 V Given current direction implies that V 1 > V 2.

16 Resistor A resistor is a passive electronic component that obeys Ohm’s Law. It has resistance R SI Unit: ohm (  ) Symbol: Impedance: I - V relationship: R i(t)i(t) v(t)v(t)

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18 Resistors Connected in Series (end-to-end) If N resistors are connected in series, with the i-th resistor having a resistance R i, then the equivalent resistance R eq is:  R1R1 R2R2 R eq = R 1 + R 2  R eq = R 1 + R 2 + R 3 R1R1 R2R2 R3R3

19 Properties of Series Resistances (DC) The amount of current I in entering one end of a series circuit is equal to the amount of current I out leaving the other end. The current is the same through each resistor in the series and is equal to I in. I in = I out = I 1 = I 2 = I 3 → I1→ I1 I in → → I2→ I2 → I3→ I3 → I out R1R1 R2R2 R3R3

20 The amount of voltage drop across each resistor in a series circuit is given by Ohm’s Law. V 1 = IR 1, V 2 = IR 2, V 3 = IR 3 By convention, the resistor terminal that the current enters is labeled “+”, and the terminal the current exits is labeled “–” + – V 1 I → → I→ I + – V 2 + – V 3 R1R1 R2R2 R3R3

21 Example 2  6  3  AB I V AB = 3 V a)Calculate the current I that flows through the resistors. b)Find the voltage drop across the 6  resistor. Assume 3 significant figures. Solution: a) Approach – Use Ohm’s law: Calculate the equivalent resistance: Calculate the current: b) Use Ohm’s law again:

22 Resistors Connected in Parallel (side-by-side) If N resistors are in parallel, with the i-th resistor having a resistance R i, then the equivalent resistance is:  R1R1 R2R2  R3R3 R1R1 R2R2

23 Properties of Parallel Resistances (DC) The amount of current I in entering one end of a parallel circuit is equal to the amount of current I out leaving the other end For parallel resistors, the voltage drop across each resistor is the same I in = I out and V 1 = V 2 = V 3 I in → → I out + V 1 – R3R3 R1R1 R2R2 + V 2 – + V 3 –

24 The amount of current through each resistor in a parallel circuit is given by Ohm’s Law. The sum of the currents through each resistor is equal to the original amount of current entering or leaving the parallel circuit I 1 + I 2 + I 3 = I I →I → → I→ I + V – R3R3 R1R1 R2R2 I1I1 I2I2 I3I3 I 1 = V / R 1 I 2 = V / R 2 I 3 = V / R 3

25 Example 2  6  3  A B I = 4 A V AB a)Find the current I x through the 2  resistor. b)What is the power dissipated by the 3  resistor? Assume 3 significant figures. Solution: a) Approach – Use Ohm’s law: Calculate the equivalent resistance: Calculate the voltage drop: b) Use power equation: Find the current: Ix Ix

26 Inductor & Capacitor An inductor is a passive component that stores energy within a magnetic field. It has inductance L SI unit: henry (H)Impedance: Symbol: A capacitor is a passive component that stores energy within an electric field. It has capacitance C SI unit: farad (F)Impedance: Symbol: non-polarized polarized +

27 DC Voltage & Current Sources + – VsVs I + – VsVs I An ideal DC current source outputs a constant current regardless of the amount of voltage across it An ideal DC voltage source outputs a constant voltage regardless of the amount of current through it IsIs V

28 Voltage & Current Dividers Voltage Divider R2R2 I R1R1 + – + – + – V0V0 V1V1 V2V2 Current Divider I + – VSVS G2G2 G1G1 I1I1 I2I2

29 Example Voltage Drop What is the voltage drop across R a ? What is the voltage drop across R b ? VbVb R a = 1 Ω VaVa + – V s = 8 V R b = 3 Ω

30 Example Resistor The voltage V g from a signal generator must be reduced by a factor of three before entering an audio amplifier. Determine the resistor values of a voltage divider that performs this task R2R2 R1R1 VgVg V out Signal Generator Audio Amplifier Peak output = 3 V Max input = 1 V

31 Example Resulting Resistance a)What is the equivalent resistance R eq (in Ω ) that is “seen” by the voltage source? b)What is the current (in A) in R 6 ? VSVS R3R3 + – R2R2 R0R0 R1R1 R 0 = 5.0  R 1 = 2.0  R 2 = 8.0  R 3 = 6.0  R 4 = 1.0  R 5 = 3.0  R 6 = 6.0  R 7 = 8.0  V S = 10.0 V R4R4 R5R5 R6R6 R7R7