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MA2009: tutorial T1

circuits fundamentalscircuits fundamentals

T1 1: batteriesT1.1: batteries

• The capacity of a car battery is usually specified in ampere‐hours. A b tt t d t 100 A h h ld b bl t l 100 A• A battery rated at, say, 100 A‐h should be able to supply 100 A for 1 hour, 50 A for 2 hours, 25 A for 4 hours, 1 A for 100 hours or any other combination yielding a product of 100 A‐hours, or any other combination yielding a product of 100 Ah.a. How many coulombs of charge should we be able to draw from a 

fully charged 100 A‐h battery?b. How many electrons?

answers:a)360x103C; b)224x1022

T1 2: batteriesT1.2: batteries

Th h l h i Fi i• The charge cycle shown in Figure is an example of a two‐rate charge. 

• The current is held constant at 50 mA for 5 h Then it is switched to 20mA for 5 h. Then it is switched to 20 mA for the next 5 h. Find:a The total charge transferred to thea. The total charge transferred to the 

battery.b. The energy transferred to the 

battery.

Hint: Recall that energy, w, is the integral of power, or P = dw/dt .gy, , g f p , /

answers:a)1260 Cb)1732.5 J

T1 3: Kirchhoff’s lawsT1.3: Kirchhoff s laws

• apply KCL to find the unknown current in the following circuitsg

a) b)

answers:a)3Ab)‐3A

T1 4: Kirchhoff’s lawsT1.4: Kirchhoff s laws

• apply KCL to find voltages v1 and v2

answers:a)v1=12Vb)v2=2V

T1 5 : Ohm’s lawT1.5 : Ohm s law

• use Ohm’s law and KCL to determine the current in the circuit

I1I2

2

answers:a) I1= 6.66Ab) I2= 3.33A

T1 6 : powerT1.6 : power

• For the circuit

– determine which components are absorbing power and which are delivering power

– is the conservation of power satisfied?• explain your answer YES (power is conserved)p y

C absorbs 25WD absorbs 30WE absorbs 20W

answers:A generates 60WB generates 15W

MP2008: tutorial T2

node‐voltage and mesh‐currentnode voltage and mesh currentmethods

T2 1: Kirchhoff’s lawsT2.1: Kirchhoff s laws

• use Kirchhoff’s Current Law (KCL) to determine the unknown currents in the circuit

• assumeI 2 AI0 = ‐2 AI1 = ‐4 AIS = 8 AVS = 12 VVS  12 V

AnswersI2 = 6A;I3 = 2A.

T2 2: node voltage methodT2.2: node voltage methodAnswers14.23V; 4.6 V; ‐5.4 V

T2 3:mesh current methodT2.3: mesh current methodAnswer23.9 V

T2 4: look carefully at the circuitT2.4: look carefully at the circuitfind simple ways to determine the current I2 for p y 2both KCL and KVL cases

II2

Answer: 1.6A

T2 5: Kirchhoff’s lawsT2.5: Kirchhoff s laws • For the circuit shown in the following Figure, determine: g g ,

a. The current I1, I2 and I3b. The power delivered by the 3 A current source and by the 12V 

lvoltage source c. The total power dissipated by the circuit. 

• Express I I and I as a function of V knowing that:

V1

• Express I1, I2 and I3 as a function of V1 knowing that:– R1 = 25 Ω ,  R2= 10 Ω,  R3= 5 Ω ,  R4= 7 Ω 

1

I3

I2

‐279 W (current source);‐6.2 W (voltage source);285.2W (tot dissipated)

Answers:I1 = 3A; I2 = ‐1.8A; I3 = ‐1.2A;PWR 

MP2008: tutorial T3

superpositionsuperpositionThevenin & Norton Equivalent

T3 1: superpositionT3.1: superpositionAnswer5.99 V

T3 2: equivalent resistorT3.2: equivalent resistor

fi d th i l t i t b th• find the equivalent resistance seen by the source and the current I in the circuit.

• Vs=12V, R0=4Ω, R1=2Ω, R2=50Ω, R3=8Ω, R4=10Ω,R5=12Ω, R6 = 6Ω . 

Answers4.76; 139 mA

T3 3: Thevenin equivalentT3.3: Thevenin equivalent

find the Thevenin equiv. resistance seen by resistor R5 in the circuit below and computer the Thevenin (open‐circuit) voltage and the Norton (short circuit) when R5 is the loadvoltage and the Norton (short‐circuit) when R5 is the load.

Answers22.92 ; ‐6.67 V;  ‐291 mA

T3 4: Thevenin & NortonT3.4: Thevenin & Norton

find the Thevenin equiv. resistance seen by resistor R5 in the circuit below and computer the Thevenin (open‐circuit) voltage and the Norton (short circuit) when R5 is the loadvoltage and the Norton (short‐circuit) when R5 is the load.

Answers:52 ; 12.8 V; 246 mA

MP2008: tutorial T4

power transferpower transferenergy storagesignal measures

T4 1:max power transferT4.1:max power transferAnswers:a)8; 4.5W; 50%b)600; 510.4mW; 50%

T4 2: energy storageT4.2: energy storageAnswers:11.76J; 0J; 0.18J; 17.65J

T4 3: capacitive currentsT4.3: capacitive currents

T4 4: rms and averageT4.4: rms and average

• Find the ratio between the average and rmsvalues of the waveform in the figureg

voltage1V

voltage

time2s 4s

9V‐9V

Answer:‐0.625

MP2008: tutorial T5

frequency responsefrequency response

T5 1: complex numbersT5.1: complex numbers

• A complex number Z can be represented in various ways. Typically y yp y– Cartesian: Polar:– Polar:

• determine the Polar and Cartesian forms for the following complex numbers

T5 2: cutoff frequencyT5.2: cutoff frequency53kHz

Answer:high‐passgain 2/3 (or ‐3.5dB)

T5 3: RLC filtersT5.3: RLC filters

(a) (b)

(c) (d)

c)LPd)LP

Answera)HPb)HP

MP2008: tutorial T6

transient analysis fortransient analysis for switched, 1st order circuits

T6 1: 1st order circuitsT6.1: 1st order circuits 

R1 = 5kR2 = 7kR3 = 3kL   =  0.9mHVs = 10V

T6 2: 1st order circuitsT6.2: 1st order circuits

T6 3: initial conditionsT6.3: initial conditionsDetermine the initial and final conditions for the following circuitsDetermine the initial and final conditions for the following circuits, assuming a steady‐state condition  at time t=0‐

T6 4: initial conditionsT6.4: initial conditionsAnswer35.4 ms

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