Chapter 1: From Physics to Electric · PDF file · 2 days agoChapter 1 From physics to electric circuits I-15 Problem 1.15. A copper wire (AWG #24) ... Chapter 1 From physics to electric

Chapter 1 From physics to electric circuits

I-1

Chapter 1: From Physics to Electric Circuits

1.1 Electrostatics of conductors 1.1.2 Electric potential and electric voltage 1.1.3 Electric voltage versus ground 1.1.4 Equipotential conductors Problem 1.1. Determine voltage (or potential) ABV and BAV (show units) given that the electric field between points A and B is uniform and has the value of 5 V/m. Point A has coordinates (0, 0); point B has coordinates (1, 1).

Solution: The integral along the path is the product of the field, path length, and the factor of ±1 (if the path direction and the E-field direction coincide, this factor is one; in the opposite case it is -1). Therefore,

25ABV V;

25BAV V.


I-2

Problem 1.2. Determine voltages ABV , BDV , and BCV given that the electric field shown

in the figure that follows is uniform and has the value of (A) 10 V/m; (B) 50 V/m; (C) 500 V/m.

Solution: (the original exam solutions are given; note problem linearity!)


I-3

Problem 1.3. Assume that the electric field along a line of force A A has the value 1×l V/m where m10 l is the distance along the line. Find voltage (or potential) AAV . Solution:

V5.00

1

ldlldlVA

A

AA


I-4

Problem 1.4. The electric potential versus ground is given in Cartesian coordinates by zyxrV )(

[V]. Determine the corresponding electric field everywhere in space.

Solution: After differentiation, we obtain

V/m1V/m,1,V/m1 zyx EEE .


I-5

Problem 1.5. Figure that follows shows the electric potential distribution across the semiconductor pn-junction of a Si diode. What kinetic energy should the positive charge (a hole) have in order to climb the potential hill from anode to cathode given that the hill “height” (or the built-in voltage of the pn-junction) is V7.0biV ? The hole charge is

the opposite of the electron charge. Express your result in joules.

Solution: Since the decrease in the kinetic energy is equivalent to the increase in the potential energy, one has

19101214.1 bikin VqE J.


I-6

Problem 1.6. Is the electric field shown in the figure that follows conservative? Justify your answer.

Solution: It is not conservative since the value of the potential depends in the integration path. Two paths shown in the figure that follows will give us two different potential values.


I-7

Problem 1.7 List the conditions for voltage and electric field used in electrostatic problems. Solution:

1. The electric field everywhere within the conductor is zero, 0E

. 2. The volumetric charge density everywhere within the conductor is zero. 3. The surface charge density is not zero.

4. The tangential component of the electric field, tE

, is zero on the entire surface of

the conductor. 5. The potential or voltage remains the same for any point of this surface. The

conductor surface is the equipotential surface.


I-8

Problem 1.8. Figure that follows shows a 345 kV power tower used in MA, USA – front view. It also depict electric potential/voltage and electric field distributions in space.

A. Determine which figure corresponds to the electric potential and which – to the magnitude of the electric field.

B. Provide a detailed justification of your answer.

Solution: Figure b) is the electric field distribution and figure c) is the voltage distribution. Reason: the steel bar should be at zero voltage/potential.


I-9

Problem 1.9. Figure that follows shows some isolated conductors. Determine the following voltage differences: CBACAB VVV ,, .

Solution: We apply the conditions of electrostatics and obtain:

.V5.2

,V0

,V5.2

CB

AC

AB

V

V

V


I-10

Problem 1.10. In your circuit, two wires connected to a 10 V voltage supply happen to be very close to each other at a certain location; they are separated by 2 mm.

A. What is the voltage between the wires at this location? B. What is approximately the electric field strength at this location?

Solution (A): 10 V Solution (B): ~5000 V/m.


I-11

Problem 1.11. Is the figure that follows correct? Black curves indicate metal conductors. Why yes or why not?

Solution: The figure is indeed correct. The voltage difference between two conductors is the same along the conductors.


I-12

1.2 Steady-state current flow and magnetostatics 1.2.1 Electric current 1.2.2 Difference between current flow model and electrostatics 1.2.3 Physical model of an electric circuit Problem 1.12. List the conditions for voltage and electric field used in the steady-state electric current problems. Solution:

1. Everywhere on the conductor surface (but not on the electrode surface), the current density component perpendicular to the surface is zero. In other words, no current can flow from the conductor into air.

2. On the surface of electrodes, either the voltage or current may be given. 3. Electric current can flow through the electrode surface.


I-13

Problem 1.13. Figure that follows shows the lines of force and equipotential lines for a DC current flow in a conductor due to two electrodes. List all mistakes of this drawing.

Solution:

1. Lines of force and equipotential lines are not perpendicular to each other; 2. Equipotential lines cannot form closed loops in the homogeneous space; 3. Equipotential lines are not perpendicular to the conductor-air interface; 4. Lines of force cannot hit the conductor surface except for the electrodes (current

cannot flow into the empty space).


I-14

Problem 1.14. An AWG #10 (American Wire Gauge) aluminum wire has the conductivity of 4.0×107 S/m and the diameter of 2.58826 mm. Determine the total current in the wire when the electric field inside the wire is

A. 0.001 V/m B. 0.005 V/m

Solution: The total current is given by

ErI 2 Therefore: Case A: I = 0.21 A Case B: I = 1.05 A Note: A MATLAB script that follows solves the problem: D = 2.58826e-3; R = D/2; A = pi*R^2; sigma = 4e7; E = 1e-3; I = A*sigma*E


I-15

Problem 1.15. A copper wire (AWG #24) in the form of a coil with the radius of 0.1m and 1000 turns is subject to applied voltage of 5 V. Determine the total current in the wire if its diameter is 0.51054 mm; the copper conductivity is 5.8×107 S/m. Solution: The electric field within the wire is given by (R is the coil radius and N is the number of turns)

)2/( RNVE The current in the wire is given by

ErI 2

Therefore, I = 0.0945 A Note: This problem does use the resistance concept explicitly; it operates using the electric field. Note: A MATLAB script that follows solves the problem: D = 0.51054e-3; R = D/2; A = pi*R^2; sigma = 5.8e7; V = 5; E = V/(1000*2*pi*0.1) I = A*sigma*E


I-16

Problem 1.16. Figure that follows shows a conducting cylinder of radius cm 1R , length cm 5L , and conductivity S/m 0.11 in air. Two electrodes are attached on both cylinder sides, the electrode radius is exactly the cylinder radius. Electrode voltages are exactly ±1V.

A. Determine and sketch to scale the electric potential everywhere inside the cylinder and on its surface.

B. Determine and sketch to scale the electric field everywhere inside the cylinder; C. Attempt to sketch the electric potential distribution outside the cylinder. D. Repeat tasks A and B when the voltage electrodes are replaced by current

electrodes with the applied electric current density of ±1 A/m2. Solution:


I-17

1.2.4 Magnetostatics/Ampere’s law 1.2.5 Origin of electric power transfer Problem 1.17. A DC magnetic field of 1000 A/m is measured between two parallel wires of an electric circuit separated by 0.5 m. What is the circuit current? Solution: The contributions of two wires add up; therefore,

15712

2 rHIr

IH

A


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Problem 1.18. Two conductors running from the source to the load are two parallel 0.5 cm wide thin plates of infinite conductivity. The (vertical) electric field between the plates is 100 V/m, the (horizontal) magnetic field between the plates is 100 A/m. The load power is 0.1 W. Assuming no field fringing determine

A. Plate separation; B. Load voltage; C. Load current.

Solution (A): The plate separation, h, is found from the equality for two powers: the load power and the power transferred by the fields between the plates, that is

2105.0

1.0105.02

2

EH

PhPEHh mm.

Solution (B):

2.0 EhV V Solution (C):

5.0/ VPI A


I-19

Problem 1.19. Repeat the previous problem when the electric field between the plate electrodes increases by the factor of two but the magnetic field decreases by the factor of two. Solution (A): The plate separation, h, is found from the equality for two powers: the load power and the power transferred by the fields between the plates, that is

2105.0

1.0105.02

2

EH

PhPEHh mm.

Solution (B):

4.0 EhV V Solution (C):

25.0/ VPI A


I-20

1.3 Hydraulic and fluid mechanics analogies Problem 1.20. For the hydraulic setup shown in the figure, draw its electrical counterpart (an electric circuit) using the circuit symbols.

Solution: A circuit with the current power source (replacing a water pump of constant speed) and a resistance (replacing a water filter); not grounded:


I-21

Problem 1.21 For the hydraulic setup shown in the figure, present its electrical counterpart (an electric circuit). Note a connection to a large reservoir with atmospheric pressure.

Solution: A circuit with the voltage power source (replacing a water pump of constant torque) and a resistance (replacing a water filter); grounded:

Chapter 2

II-1

Chapter 2: Major Circuit Elements Section 2.1 Resistance – linear passive circuit element 2.1.2 Resistance 2.1.3 -i Characteristic of the resistance. Open and short circuits 2.1.4 Power delivered to the resistance 2.1.5 Finding resistance of ohmic conductors Problem 2.1. Plot -i characteristics of the following resistances: (A) k8 ; (B) k2 ; (C) k1 ; (D) 500 . Clearly label each characteristic.

Solution: The solution is shown in the figure that follows. It is interesting to observe the counterclockwise rotation of the -i characteristic for the resistance when the resistance value continuously decreases.

Chapter 2

II-2

Problem 2.2. Plot -i characteristics of the following resistances: (A) 667.1 ; (B) open circuit on the same graph.

Solution: (the original exam solution is given)

Chapter 2

II-3

Problem 2.3. Plot -i characteristics of the following resistances: (A) 5.2 ; (B) short circuit on the same graph.


Chapter 2

II-4

Problem 2.4. Given -i characteristics of a resistance determine the corresponding conductance. Show units.

Solution (A):

1m3 G or 3 mS. Solution (B):

10 G or 0 S.

Chapter 2

II-5

Problem 2.5. An incandescent energy-saving light bulb (“soft white”) from General Electric is rated to have the wattage of 57 W when the applied AC voltage is 120 V rms (root mean square). This means that the corresponding DC voltage providing the same power to the load is exactly 120 V. When the bulb is modeled as a resistance, what is the equivalent resistance value?

Solution: The (equivalent) DC voltage across the equivalent resistance (the light bulb) is 120 V.

Therefore, 25357

120222

P

VR

R

VP .

Chapter 2

II-6

Problem 2.6. Power absorbed by a resistor from the ECE laboratory kit is 0.2 W. Plot the -i characteristics of the corresponding resistance to scale given that the DC voltage across the resistor was 10 V.

Solution: The solution is shown in the figure that follows.

Chapter 2

II-7

Problem 2.7. The number of free electrons in copper per unit volume is

328

m

11046.8 n . The charge of the electron is −1.60218×10−19 C. A copper wire of

cross-section 0.25 mm² is used to conduct 1A of electric current. A. Sketch the wire, the current direction, and the direction of electron motion. B. How many Coulombs per one second is transported through the conductor? C. How fast do the electrons really move? In other words, what is the average electron

velocity? Solution:

1. An electric current is defined as the net flow of positive charges in a conductor, and is therefore opposite to the direction of electron motion. The figure follows:

2. A copper wire conducts 1A of electric current. By definition, s1

C1A1 . It means

than one Coulomb of charge is transported per one second.

3. We use the equation AI , and solve for : I and A are given:

A = 0.25 mm² = 26 m1025.0 I = 1A nq ; n and q are given:

n = 3

28

m

11046.8

q = C1060218.1 19 We now solve for :

s

mm3.0

s

m1095.2

C1060218.1m

11046.8m1025.0

A1 4

193

2826

A

I

The negative sign indicates that the electrons move in the opposite direction of the electric current. The electron motion is thus very slow. However, the transported charge may be large.

Chapter 2

II-8

Problem 2.8. Repeat the above problem when the conductor’s cross-section is increased to 5 mm². Solution: 1. The solution is the same as in Problem 2.5. 2. The solution is the same as in Problem 2.5. 3. The average velocity decreases by the factor of 20 compared to Problem 2.5 since the cross-section increases by the factor of 20, but the current remains the same:

s

μm15

s

m1048.1

C1060218.1m

11046.8m105.0

A1 5

193

2825

A

I

Chapter 2

II-9

Problem 2.9. A copper wire having a length of 1000 ft and a diameter of 2.58826mm is used to conduct an electric current of 5 A.

A. What is wire's total resistance? Compare your answer to the corresponding result of Table 2.2.

B. What is the power loss in the wire?

Solution (A):

We use the definition to find the resistance: A

L

A

LR

. We must convert all units to

the SI system first: Given: L = 1000 ft =304.8m A = (2.58826mm/2 )2 = 5.26 mm2 =5.2610-6 m2 The material conductivity, , of copper is found from Table 2.1:

= 5.8 × 107 S/m (S stands for the unit Siemens;

1

1S1 )

Now,

999.0

m

S105.8m1052.6

m8.304

727-A

LR

Table 2.2 gives the same result. Solution (B): We can now solve for power loss as follows:

W25999.0A)5( 22 RIP The entire power absorbed by the wire is the power loss. This power is transformed into heat.

Chapter 2

II-10

Problem 2.10. A. A copper wire having a length of 100 m and a cross-section of 0.5 mm2 is used to

conduct an electric current of 5 A. What is the power loss in the wire? Into what is this power loss transformed?

B. Solve task A when the wire cross-section is increased to 2.5 mm2. Solution (A):

We use the Ohm’s law to find the resistance: A

L

A

LR

Given: L = 100 m A = 0.5 mm2 = 5 × 10-7 m2

The material conductivity, , of copper is found from Table 2.1:

= 5.8 × 107 S/m (S stands for the unit Siemens;

1

1S1 )

Now,

45.329

100

m

S105.8m105

m100

727-A

LR

We can now solve for the absorbed power as follows:

W21.8629

100A)5( 22 RIP

The entire power absorbed by the wire is the power loss. This power is transformed into heat. Solution (B): The same method as in (A) applies, but the wire resistance becomes five times less. Hence, the power loss becomes five times less:

W24.1729

20A)5( 22 RIP

Chapter 2

II-11

Problem 2.11. Determine the total resistance of the following conductors (show units): A. A cylindrical silver rod of radius 0.1mm, length 100mm, and conductivity 6.1×107

S/m. B. A square graphite bar with the side of 1mm, length 100mm, and conductivity

3.0×104 S/m. C. A semiconductor doped Si wafer with the thickness of 525m. Carrier mobility is

15.0 m2/(Vs). Carrier concentration is 323 m10 n . Carrier charge is 1.6×10-

19 C. The resistance is measured between two circular electrodes with the radius of 1mm each, which are attached on the opposite sides of the wafer. Assume uniform current flow between the electrodes.

Solution:

a) 052.0R . b) 3.3R .

c) 0695.0,A

LRnq

.

Chapter 2

II-12

Problem 2.12. A setup prepared for a basic wireless power-transfer experiment utilizes a square multi-turn loop schematically shown in the figure, but with 40 full turns. A #22 gauge copper wire with the diameter of 0.645mm is used. Total loop resistance, R, is needed. Please, assist in finding the loop resistance (show units).

Solution:

66.12A

LR

Chapter 2

II-13

Problem 2.13. Estimate resistance, nR (show units), of the n-side of a Si pn-junction

diode in the figure that follows. We model the n-side by a Si bar having the following parameters:

i. length of L = 0.0005 cm = 5 m; ii. cross-section of A=0.01 cm0.01 cm=110-4 cm2; iii. uniform electron concentration (carrier concentration) of 317 cm10 n . This value

is typical for a Si diode pn-junction. Carrier mobility is 1450n cm2/(Vs).


22.0,A

LRnq n

Chapter 2

II-14

Problem 2.14. Estimate resistance, pR (show units), of the p-side of a Si pn-junction

diode in the figure that follows. We model the p-side by a Si bar having the following parameters:

i. length of L = 0.0005 cm = 5 m; ii. cross-section of A=0.01 cm0.01 cm=110-4 cm2; iii. uniform hole concentration (carrier concentration) of 317 cm10 n . This value is

typical for a Si diode pn-junction. Carrier mobility is 500n cm2/(Vs).


625.0,A

LRnq n

Chapter 2

II-15

Problem 2.15. A cross-section of the most popular NMOS transistor is shown in the figure that follows. The transistor has three terminals (metal contacts): drain (with voltage 0DSV vs. source), gate (with voltage 0GSV vs. source), and source itself

(grounded). The source is also connected to a metal conductor on the other side of the semiconductor body. Accordingly, there are two types of the electric field within the semiconductor body: the horizontal field created by GSV , and the vertical field created by

DSV . The horizontal field fills a conducting channel between the drain and the source

with charge carriers, but has no effect on the vertical charge motion. The resulting carrier concentration in the channel is given by constVconstNVVNn ThThGS ,,0)( .

The individual carrier charge is q. Given the channel cross-section A, the carrier mobility , and the channel length L, determine transistor current DI and transistor resistance

(drain-to-source resistance) DSR . Express both results in terms of quantities listed above

including GSV and ThV .


Chapter 2

II-16

Chapter 2

II-17

2.1.6 Application example: power loss in transmission wires and cables Problem 2.16. An AWG 0000 aluminum transmission grid cable has the wire diameter of 11.68 mm and the area of 107 mm2. The conductivity of aluminum is 4.0107 S/m. The total cable length (two cables must run to a load) is 100km. The system delivers

MW 10 of DC power to a load. Determine the power loss in the cable (show units) when load voltage and current are given by:

1. V =200 kV and I =50 A 2. V =100 kV and I =100 A 3. V =50 kV and I =200 A

Why do you think the high-voltage power transmission is important in power electronics?

Solution : We use the resistance definition to find the total cable resistance and obtain 23.33Ω. The power loss in the cable is given by 2RIPloss . We obtain

1. lossP = 58 kW or 0.6% of load lower of 10 MW

2. lossP = 233 kW or 2.3% of load power of 10 MW

3. lossP = 933 kW or 9.3% of load power 10 MW

The advantage of a high voltage power transmission becomes obvious – a better transmission efficiency when transmitting power over large distances.

Chapter 2

II-18

Problem 2.17. Solve the previous problem when the total cable length (two cables must run to a load) is increased to 200km. Solution: We use the resistance definition to find the total cable resistance and obtain 46.66Ω. The power loss in the cable is given by 2RIPloss . We obtain

1. lossP = 116 kW or 1.2% of load lower of 10 MW

2. lossP = 466 kW or 4.7% of load power of 10 MW

3. lossP = 1866 kW or 18.7% of load power 10 MW

Chapter 2

II-19

Problem 2.18. An AWG 00 aluminum transmission grid cable has the wire diameter of 9.266 mm. The conductivity of aluminum is 4.0107 S/m. A power transmission system that uses this cable is shown in the figure that follows. The load power is 1 MW. Determine the minimum necessary load voltage V that guarantees us a 1% relative power loss in the cables.

Solution: We use the resistance definition to find the total cable resistance and obtain 74.15Ω. The power loss in the cable is given by 2RIPloss . The ratio of this power to the load power

is given by 0.01; this allows us to find the load current:

A6.11kW10

01.0MW1

01.0 22

IR

IRI

P

P

Load

loss

Therefore, the minimum load voltage (the load voltage virtually coincides with the line voltage) should be given by

86.2kVMW1

I

V

Chapter 2

II-20

Problem 2.19. An AC-direct micro hydropower system is illustrated in the figure that follows.

Reprinted from Micro-Hydropower Systems Canada 2004, ISBN 0-662-35880-5. The system uses a single phase induction generator with the rms voltage (equivalent DC voltage) of 240V. The system serves four small houses, each connected to the generator via a separate transmission line with the same length of 1000 m. Each line uses AWG#10 aluminum wire with the diameter of 2.59 mm. The conductivity of aluminum is 4.0107 S/m. The house load is an electric range with the resistance of 20 . Determine total power delivered by the generator, totalP , total power loss in the transmission lines, lossP ,

and total useful power, usefulP (show units).


Chapter 2

II-21

Chapter 2

II-22

2.1.7 Physical component – resistor Problem 2.20. A leaded resistor has color bands in the following sequence: brown, black, red, gold. What is the resistor value? Solution: The color band values are shown in Table 2.3. The first two bands make up the base resistance (the first band is the tens digit; the second band is the ones digit), the third band is the multiplier value, and the fourth band is the tolerance. The color values for the bands in this resister are as follows: brown = 1 black = 0 red = 2 gold = 5% tolerance Thus: tolerance%5 with ,k1100010)0101( 2 R

Chapter 2

II-23

Problem 2.21. Potentiometer operation may be schematically explained as moving a sliding contact #2 in the figure that follows along a uniform conducting rod with the total resistance of 20 k. Determine resistance between terminals 1 and 2 as well as between terminals 2 and 3 of the potentiometer, when the sliding contact is at one fifth of the rod length.

Solution: 16 k and 4 k, respectively.

Chapter 2

II-24

Section 2.2 Nonlinear passive circuit elements 2.2.2 Nonlinear passive circuit elements 2.2.3 Static resistance of a nonlinear element 2.2.4 Dynamic (small-signal) resistance of a nonlinear element 2.2.5 Electronic switch Problem 2.22. A nonlinear passive circuit element – the ideal diode – is characterized

by the the -i characteristic in the form

1exp

TS V

VII with A101 13SI and

7.25TV mV. Find the static diode resistance 0R and the diode current 0I when

A. 0V =0.40 V;

B. 0V =0.50 V;

C. 0V =0.55 V;

D. 0V =0.60 V.

Solution (A, B, C, D): The static resistance is given by

1exp

)(

TS V

VI

VVR

The particular values are:

A. 0V =0.40 V μA575.0,k696 00 IR ;

B. 0V =0.50 V μA1.28,k8.17 00 IR ;

C. 0V =0.55 V μA197,k79.2 00 IR ;

D. 0V =0.60 V mA4.1,436 00 IR .

Chapter 2

II-25

Problem 2.23. Find the dynamic (small-signal) resistance r of a nonlinear passive circuit element – the ideal diode – when the operating DC point 00 , IV is given by the solutions

to the previous problem. Consider all four cases. Solution (A, B, C, D): The dynamic resistance is given by

S

T

II

V

dI

dVIr

)( ,

1exp

TS V

VII


A. 0V =0.40 V μA575.0,k696 00 IR r = 44.7 k;

B. 0V =0.50 V μA1.28,k8.17 00 IR r = 913 ;

C. 0V =0.55 V μA197,k79.2 00 IR r = 131 ;

D. 0V =0.60 V mA4.1,436 00 IR r = 18.7 k.

Note: The solution to the problem is given by a simple MATLAB script VT = 0.0257 IS = 1.0e-13; V =0.60; I = IS*exp(V/VT) R = V/I r = VT/(I + IS)

Chapter 2

II-26

Problem 2.24. A nonlinear passive circuit element is characterized by the -i

characteristic in the form 2)/(1

/

S

SS

VV

VVII

with A1SI and V1SV . Plot the -i

characteristic to scale. After that, find the static element resistance 0R , the element

current 0I , and the corresponding dynamic element resistance r when

A. 0V =0.1 V;

B. 0V =1.0 V;

C. 0V =5.0 V.

Solution: The plot is shown in the figure that follows.

The solutions for static and dynamic resistances are equivalent to the calculation of a known function and its derivative at the operating point. The static resistance is given by

2)/(1)( SS

S VVI

VVR


A. 0V =0.1 V A10.0,0.1 00 IR ;

B. 0V =1.0 V A71.0,4.1 00 IR ;

C. 0V =5.0 V A98.0,1.5 00 IR .

Chapter 2

II-27

. The dynamic resistance is given by

22 )/(1

//

)/(1

/

S

SS

S

SS

II

IIVV

VV

VVII

2/32 ))/(1(

1)(

SS

S

III

V

dI

dVIr


A. 0V =0.1 V A10.0,0.1 00 IR r = 1.0 ;

B. 0V =1.0 V A71.0,4.1 00 IR r = 2.8 ;

C. 0V =5.0 V A98.0,1.5 00 IR r = 133 .

Note: The problem may be solved with the help of the MATLAB script: clear all VS = 1 IS = 1; % V = [-5:0.1:5]; % I = IS*(V/VS)./sqrt(1 + (V/VS).^2); % plot(V, I); grid on V = 5.0; I = IS*(V/VS)./sqrt(1 + (V/VS).^2) R = V/I r = VS/IS/(1-(I/IS)^2)^1.5

Chapter 2

II-28

Problem 2.25. Repeat the previous problem when A5.0SI and V5.0SV . All other

parameters remain the same. Consider the following DC operating points A. 0V =0.05 V;

B. 0V =0.50 V;

C. 0V =2.50 V.

Solution: The plot is shown in the figure that follows.

The solutions for static and dynamic resistances are equivalent to the calculation of a known function and its derivative at the operating point. The static resistance is given by

2)/(1)( SS

S VVI

VVR


D. 0V =0.05 V A05.0,0.1 00 IR ;

E. 0V =0.50 V A35.0,4.1 00 IR ;

F. 0V =2.50 V A49.0,1.5 00 IR .

. The dynamic resistance is given by

22 )/(1

//

)/(1

/

S

SS

S

SS

II

IIVV

VV

VVII

2/32 ))/(1(

1)(

SS

S

III

V

dI

dVIr


D. 0V =0.05 V A05.0,0.1 00 IR r = 1.0 ;

Chapter 2

II-29

E. 0V =0.50 V A35.0,4.1 00 IR r = 2.8 ;

F. 0V =2.50 V A49.0,1.5 00 IR r = 133 .

Note: It is important to emphasize that the solutions for the resistances do not change compared to the previous problem. This fact underscores the importance of voltage and current scaling considerations, which remain valid in the nonlinear case too. Note: The problem may be solved with the help of the MATLAB script: clear all; clc VS = 0.5; IS = 0.5; % V = [-5:0.1:5]; % I = IS*(V/VS)./sqrt(1 + (V/VS).^2); % plot(V, I); grid on V = 2.5; I = IS*(V/VS)./sqrt(1 + (V/VS).^2) R = V/I r = VS/IS/(1-(I/IS)^2)^1.5

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Chapter 1: From Physics to Electric · PDF file · 2 days agoChapter 1 From physics to electric circuits I-15 Problem 1.15. A copper wire (AWG #24) ... Chapter 1 From physics to electric