Vietnamese German University, Ho Chi Minh City Prof. Dr.-Ing. habil. Joachim Laemmel

1. How many electrons are at the electrodes of a capacitor if the electric charge is

Exercises in Fundamentals of Electrical Engineering

Q = -0.1C ? solution: z = 6.25x1017

2. Which electric field strength exists in the field of a capacitor with plate electrodes at a voltage between the electrodes of U = 220V and a distance of the plates of 5cm or 0.01mm ?

solution: E = 44V/cm and E = 220kV/cm

3. A plate capacitor with an area of the electrodes of A = 78 cm² and a distance between the plates of d = 0.4mm takes over a charge of Q = 6.63⋅10-7As at a voltage of U = 600V. Calculate the relative permittivity of the insulating material between the electrodes. solution: εr = 6.4

4. Calculate the total capacity of the following circuits:

solution: C41 = 0,125nF

solution: C42 = 5,75nF

solution: C43 = 2nF

solution: C44 = 1/3nF

VGU/2012/E1/Lae

solution: C45 = 27,3pF

C46 = 5,2µF

C47 = 720pF

5. Two capacitors in series connection both with a capacity of C = 3nF are charged at a voltage of U = 30V. Calculate the charge of both capacitors.

solution: Q1 = Q2 = 45x10-9As

6. A capacitor with two plate electrodes and air between is connected at a voltage of U = 220V. a) Calculate the charge Q1 on the electrodes. b) Now we disconnect the voltage source. How charge Q2 and voltage U2 are changed

if clear water (εr = 80.8) is brought between the electrodes? c) What happens if the voltage source in case b) remains connected? solution: a) Q1 = 22x10-9As; b) Q1 = Q2; U2 = 2.75V; c) Q3 = 17.8x10-7As

7. A capacitor with air as dielectric is charged at a voltage of U = 10kV. A charge results of Q = 2x10-8As. After disconnecting the voltage source the capacitor is casted with paraffin (εr = 2) including the space between the electrodes.

a) What is the value of the capacitance before casting? b) Calculate the charge, the voltage and the capacitance behind casting. solution: a) C1 = 2pF; b) Q1 = Q2 = 2x10-8As; U2 = 5kV; C2 = 4pF

8.. A capacitor with a homogeneous field in two layers is given. The applied voltage is U = 10kV. The insulating material of the first layer is air (d1 = 9.5mm, εr1 = 1). The

material of the second layer is hard paper (d2 = 0.5mm, εr2 = 5). In which way the voltage will be distributed on the two layers? Calculate the electric

field strengths (E1 and E2) inside the layers. solution: U1 = 9,896kV; U2 = 0,104kV; E1 = 10.4kV/cm; E2 = 2.1kV/cm

VGU/2012/E2/Lae

9. Calculate the number of electrons flowing through a cross section of a conductor per second, if the current is 1A. solution: z = 62.5x1017 10. A coil has 5000 windings. The diameter of the windings is d = 6.5cm. Calculate the electric field strength E inside the conductor, if the voltage between both ends of the coil is U = 7.9V . solution: E = 7.74mV/m 11. By means of a copper wire (κ = 56,2⋅106S/m, l = 1m und A = 1mm²) an 12-V accumulator is short-circuited. Calculate the current trough the conductor. (The allowed current for copper wire at this cross section is I = 20A!) solution: I = 675A 12. A cable with two wires and a length of l = 1000m has a short circuit between the wires on its length. By means of a measuring bridge we measure between the beginning of this two wires a resistance of R = 39,22Ω. Calculate the distance between the place of measurment and the short circuit, if the diameter of the wires is d = 0,8mm and the conductivity of copper κCu = 56,3m/Ωmm². solution: l = 555m 13. The ui-characteristic of a diode is given by the following values U / V 0.4 0.6 0.8 1.0 1.1 1.2 I / A 0.115 0.1 0.8 5.0 10.1 17.0 Draw the ui-characteristic and determine the d.c. resistance and the a.c. resistance in the point (1.0V/5A). solution: R = 0,2Ω, r = 30mΩ

14. Calculate the resistance of a conductor rope from aluminium with a cross section of A = 120mm2, ρ = 0,02857Ωmm²/m and a length of l = 1km. Determine the losses per kilometer, if the current flow is I = 390A. solution: R = 0,238Ω, PV = 36,2kW 15. Calculate the resistance of the series connection of the following circuits: a) R1 = R2 = 100Ω b) R1 = 1Ω, R2 = 2Ω, R3 = 100Ω. solution: a) R = 200Ω b) R = 103Ω 16. Calculate the resistance of the parallel connection of the following circuits: a) R1 = R2 = 100Ω, b) R1 = 1Ω, R2 = 100Ω, c) R1 = 200Ω, R2 = 200Ω, R3 = 100Ω, d) R1 = R2 = R3 = R4 = 4Ω. solution: a) R = 50Ω b) R ≈ 1Ω, c) R = 50Ω d) R = 1Ω

VGU/2012/E3/Lae

17. Calculate the resistance of the following circuits: a) b)

solution: R11 = 6Ω solution: R12 = 42Ω c) d)

solution: R13 = 5,5Ω solution: R14 = 8,6Ω

18. Calculate the value of the resistance RAB. solution: RAB = 8kΩ

19. Determine by means of the voltage- or/and the current divider rule the following quantities a) U2n b) I1n c) I2n and d) I4n if R1 = 6Ω, R2 = R3 = 4Ω, R4 = 5Ω, U1 = 10V. 1.) 2.) 3. )

VGU/2012/E4/Lae

solution: a) U21 = 4V; U22 = 2.5V = U23 b) I11 = 1A; I12 = 1.25A = I13 c) I21 = I11 = 1A; I22 = I12/2 = 0.625A = I23 d) I4 = 2A 20. Calculate the voltage U2, if U1 = 10V. R1 24= Ω , R R R2 3 4

12= = = Ω solution: U2 6= V

21. Which value have the Resistor R, if U U2 10 2= , is true? solution: R = 2 57, Ω

22. Calculate the current trough the resistor Rx, if the total current I of the circuit is I = 10A. The parameter of the resistors are R1 = 24Ω; R2 =19Ω and Rx = 72Ω. solution: Ia = Ib = 2,5A 23. The voltage drop over the resistor R1 is UR1 = 100V. a) Calculate the current I2 trough R2. b) How much is the power loss in R3? R1 = 10kΩ; R2 = 12kΩ; R3 = 36kΩ solution: a) I2 7 5= , mA, b) PR 3 mW= 225 24. a) A 220-V light bulb has a power of P = 60W. Calculate the current. b) Which power is lost in the bulb, if the voltage is increasing by 20%? solution: a) I = 273mA b) P = 86W

VGU/2012/E5/Lae

25. Which is the power loss in the beside shown circuit? a) for R R1 2 10= = Ω and R3 15= Ω b) for U = 100V and the resistors of a) c) for R R1 2 1= = kΩ, R3 1 5= , kΩ solution: a) P = 20W b) P = 500W c) P = 200mW

26a) Formulate the node and mesh equations for the shown circuits.

solution: K1: 0 1 2 6= + − −I I I K 2: 0 2 3 5= + + +I I I K 3: 0 1 3 4= − − −I I I MI: − = − − +U I R I R I Rq1 1 1 2 2 3 3 MII: + = − + +U I R I R I Rq2 3 3 4 4 5 5 MIII: 665522q3 RIRIRIU −−+=− 26b)

solution: K1: 0 2 3 5= − + +I I I K 2: 0 1 3 4= − − −I I I K 3: 0 4 5 6= + − −I I I MI: U U I R R I Rq2 q1− = + + −1 1 2 3 3( ) MII: U U I R I Rq3 5 5 6 6− = + −q2 MIII: U I R I R R I Rq4 = + − + −3 3 4 4 7 5 5( )

27. Calculate the open-circuit voltages UlAB and the internal resistances Ri of the following active terminal circuits a) b)

solution: UU RR RlAB

q1=+

2

1 2, R R R

R Ri = +1 2

1 2 solution: U

U RR R

UlABq1

q2=+

+2

1 2, R R R

R Ri = +1 2

1 2

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c) d)

solution: U U UlAB q1 q2= − , Ri = 0 solution: UU U R

R RlABq1 q2=+

+

( ),2

1 2 R R R

R Ri = +1 2

1 2

e) f)

solution: U UlAB q2= , Ri = 0 solution: UU RR R

UlABq1

q2=+

−2

1 2, R R R

R RRi = +

+1 2

1 23

g)

solution: UU U R

R RlABq1 q2=−

+

( ),2

1 2 R R R

R Ri = +1 2

1 2

28. Calculate the source voltage Uq and internal resistance Ri of the sketched active terminal circuit.

VGU/2012/E7/Lae

U U Uq1 q2 q3 V,= = = 10 R1 6= Ω, R2 14= Ω, R R3 4 20= = Ω, R5 15= Ω solution: Ri = 25Ω: Uq = 25V

29. Determine the current trough the resistor R2 by means of a procedure of your choose. Uq1 V= 12 5, , Uq2 V,= 20 R1 5 5= , ,kΩ R2 20= kΩ, R R k3 4 10= = Ω solution: IR 2 mA= 1

30. a) Calculate for the beside sketched combination the parameter for the active and the passive terminal circuit. b) Determine in consequence the current trough the resistor R5. Uq1 V,= 10 Uq2 V,= 20 R2 15= Ω, R3 5= Ω, R4 24= Ω, R5 12= Ω solution: a) Uqers V= 20 , Riers = 20Ω, Raers = 8Ω, b) I5 0 47= , A

31. Calculate the internal resistances of both voltage divider from the low voltage side.

a) b)

solution: a) Ria = 18Ω, b) Rib = 6Ω

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32. A DC generator supplies a motor, an electric haeter and a light equipment. a) Find the open-circuit voltage Uq related to the terminals A and B. b) Calculate the internal resistance Ri of the aktive terminal circuit. c) Give the equation for the calculation of the resistance of the passive terminal circuit. d) Determine the current of the heater.

solution: a) UU R R U R

R R Rqersqg im l qm ig

ig im l=

+ +

+ +

( ), b) R

R R RR R Riers

ig l im

ig l im=

+

+ +

( ), c) R R R

R RaersB H

B H=

+

d) I URH

H= mit U

R UR R

=+

aers qers

iers aers

33. Two generators with different internal resistances supply a motor: G1: Uq1 = 62V Ri1 = 0,5Ω; M: Rim = 0,6125Ω, G2: Uq2 = 60V Ri2 = 0,3Ω;. The internal voltage source of the motor induced in the rotor is changing with the load (speed) of the motor. In different operation points results a) Uqm = 58V, b) Uqm = 52V, c) Uqm = 44,75V. Calculate the voltage at the motor terminals UM = UAB and the currents of the generators. solution: UM = 60,1V/58,7V/57V; IG1 = 3,79A/6,6A/10A; IG1 = -0,35A/4,43A/10A

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