Dr. Gyurcsek István - vili.pmmf.huvili.pmmf.hu/~gyurcsek/PUB2GY/4.10X-LCE.pdf · Villamosságtan...

1 [email protected] 2017.07.24.

DR. GYURCSEK ISTVÁN

Capacitors and Inductors - Examples

Sources and additional materials (recommended)

Dr. Gyurcsek – Dr. Elmer: Theories in Electric Circuits, GlobeEdit, 2016, ISBN:978-3-330-71341-3

Ch. Alexander, M. Sadiku: Fundamentals of Electric Circuits, 6th Ed., McGraw Hill NY 2016, ISBN: 978-0078028229

Simonyi K.: Villamosságtan. AK Budapest 1983, ISBN:9630534134

Dr. Selmeczi K. – Schnöller A.: Villamosságtan 1. MK Budapest 2002, TK szám: 49203/I

Dr. Selmeczi K. – Schnöller A.: Villamosságtan 2. TK Budapest 2002, ISBN:9631026043


Stored Energy in Capacitors

LCE.01 – Calculate energy stored in each capoacitor.

Solution

𝑖 = 63

3 + 2 + 4= 2 𝑚𝐴

𝑣1 = 2000𝑖 = 4 𝑉 , 𝑣2 = 4000𝑖 = 8 𝑉

𝑤1 =1

2𝐶1𝑣1

2 =1

2∙ 2 ∙ 10−3 ∙ 16 = 16 𝑚𝐽

𝑤2 =1

2𝐶2𝑣2

2 =1

2∙ 4 ∙ 10−3 ∙ 64 = 128 𝑚𝐽


Stored Energy in Capacitors

LCE.02 – Calculate energy stored in each capoacitor.

Solution 𝑤1 = 810 𝜇𝐽 , 𝑤2 = 135 𝜇𝐽


Equivalent Capacitance

LCE.03 – Find the equivalent capacitance between a-b

Solution

20 ∙ 5

20 + 5= 4 𝜇𝐹 → 4 + 6 + 20 = 30 𝜇𝐹 →

30 ∙ 60

30 + 60= 20 𝜇𝐹



LCE.04 – Find the equivalent capacitance.

Solution 40 𝜇𝐹



LCE.05 – Find the voltage across each capacitor.

Solution

𝐶𝑒𝑞 =1

160 +

130 +

120

= 10 𝑚𝐹

𝑞 = 𝐶𝑒𝑞𝑣 = 10𝑚 ∙ 30 = 300 𝑚𝐶

𝑣1 =𝑞

𝐶1=300𝑚

20𝑚= 15 𝑉 𝑣2 =

𝑞

𝐶2=300𝑚

30𝑚= 10 𝑉

𝑣3 =𝑞

𝐶34=

300𝑚

40 + 20 𝑚= 5 𝑉 𝑜𝑟 …𝑣3 = 30 − 𝑣1 − 𝑣2 = 5 𝑉



LCE.06 – Find the voltage across each capacitor.

Solution 𝑣1 = 30 𝑉 , 𝑣2 = 30 𝑉 , 𝑣3 = 10 𝑉 , 𝑣4 = 20 𝑉


Inductor Circuit Examples

LCE.07 – vC = ?, iL = ?, wC = ? wL = ?

Solution

𝑖 = 𝑖𝐿 =12

1 + 5= 2 𝐴

𝑣𝐶 = 5𝑖 = 10 𝑉

𝑤𝐶 =1

2𝐶𝑣𝐶

2 =1

2∙ 1 ∙ 100 = 50 𝐽

𝑤𝐿 =1

2𝐿𝑖𝐿

2 =1

2∙ 2 ∙ 4 = 4 𝐽


Inductor Circuit Examples

LCE.08 – vC = ?, iL = ?, wC = ? wL = ?

Solution 𝑣𝑐 = 6 𝑉 , 𝑖𝐿 = 3 A , 𝑤𝐶 = 72 𝐽 , 𝑤𝐿 = 27 𝐽


Equivalent Inductance

LCE.09 – Find the equivalent inductance.

Solution 𝐿𝑒𝑞 = 4 +7 × 42

7 + 42+ 8 = 4 + 6 + 8 = 18 𝐻



LCE.10 – Find the equivalent inductance.

Solution 25 mH



LCE.11 –

Solution

𝑖 𝑡 = 4 2 − 𝑒−10𝑡 𝑚𝐴 , 𝑖2 0 = −1𝑚𝐴

𝑎) 𝑖1 0 =?

𝑏) 𝑣 𝑡 =?, 𝑣1 𝑡 =? , 𝑣2 𝑡 =?

𝑐) 𝑖1 𝑡 =?, 𝑖2 𝑡 =?

𝑎) 𝑖 0 = 4 2 − 𝑒0 = 4 𝑚𝐴 → 𝑖1 0 = 𝑖 0 − 𝑖2 0 = 4 − −1 = 5 𝑚𝐴

𝑏) 𝐿𝑒𝑞 = 2 + 4 × 12 = 2 + 3 = 5 𝐻 𝑣 𝑡 = 𝐿𝑒𝑞𝑑𝑖

𝑑𝑡= 5 ∙ 4 ∙ −1 −10 𝑒−10𝑡 = 200 𝑒−10𝑡 𝑚𝑉

𝑣1 𝑡 = 2𝑑𝑖

𝑑𝑡= 2 ∙ −4 −10 𝑒−10𝑡 = 80 𝑒−10𝑡 𝑚𝑉

𝑣 𝑡 = 𝑣1 𝑡 + 𝑣2 𝑡 → 𝑣2 𝑡 = 𝑣 𝑡 − 𝑣1 𝑡 = 120 𝑒−10𝑡 𝑚𝑉



𝑖1 𝑡 =1

4න

0

𝑡

𝑣2𝑑𝑡 + 𝑖1 0 =120

4න

0

𝑡

𝑒−10𝑡𝑑𝑡 + 5 𝑚𝐴

𝑐) 𝑖1 𝑡 =?, 𝑖2 𝑡 =?

= ቚ−3𝑒−10𝑡0

𝑡+ 5 𝑚𝐴 = −3𝑒−10𝑡 + 3 + 5 = 8 − 3𝑒−10𝑡 𝑚𝐴

𝑖2 𝑡 =1

12න

0

𝑡

𝑣2𝑑𝑡 + 𝑖2 0 =120

12න

0

𝑡

𝑒−10𝑡𝑑𝑡 − 1 𝑚𝐴

= ቚ−𝑒−10𝑡0

𝑡− 1 𝑚𝐴 = −𝑒−10𝑡 + 1 − 1 = −𝑒−10𝑡 𝑚𝐴 𝐶ℎ𝑒𝑐𝑘 → 𝑖 𝑡 = 𝑖1 𝑡 + 𝑖2(𝑡)



LCE.12 –

Solution

𝑖1 𝑡 = 0.6𝑒−2𝑡 𝐴 , 𝑖 0 = 1.4 𝐴

𝑎) 𝑖2 0 =?

𝑏) 𝑖2 𝑡 =?, 𝑖 𝑡 =?

𝑐) 𝑣1 𝑡 =?, 𝑣2 𝑡 =?, 𝑣 𝑡 =?

𝑎) 𝑖2 0 = 0.8 𝐴

𝑏) 𝑖2 𝑡 = −0.4 + 1.2𝑒−2𝑡 𝐴, 𝑖 𝑡 = −0.4 + 1.8𝑒−2𝑡 𝐴

𝑐) 𝑣1 𝑡 = −36𝑒−2𝑡 𝑉, 𝑣2 𝑡 = −7.2𝑒−2𝑡 𝑉, 𝑣 𝑡 = −28.8𝑒−2𝑡 𝑉


Integrator Example

LCE.13

Solution (summing intergaror)

𝑣1 = 10 cos 2𝑡 𝑚𝑉, 𝑣2 = 0.5𝑡 𝑚𝑉, 𝑣𝐶 0 = 0 𝑉

𝑣0 = −1

𝑅1𝐶න𝑣1𝑑𝑡 −

1

𝑅2𝐶න𝑣2𝑑𝑡 = −

1

3 ∙ 106 ∙ 2 ∙ 10−6න

0

𝑡

10 cos 2𝑡 𝑑𝑡 −1

100 ∙ 103 ∙ 2 ∙ 10−6න

0

𝑡

0.5𝑡 𝑑𝑡

= −1

6

10

2sin 2𝑡 −

1

0.2

0.5𝑡2

2= −0.833 sin 2𝑡 − 1.25𝑡2 𝑚𝑉

𝑣𝑜 =?


𝑣𝑖

Differentiator Example

LCE.14 Solution (differentiator circuit)

𝑣𝑜 0 = 0 𝑉

𝑣𝑖(𝑉)

𝑅𝐶 = 5 ∙ 10−3 ∙ 2 ∙ 10−6 = 1𝑚𝑠

𝑣𝑖 = ቊ2000𝑡 0 < 𝑡 < 2 𝑚𝑠

8 − 2000𝑡 2 < 𝑡 < 4 𝑚𝑠

𝑣𝑜 = −𝑅𝐶𝑑𝑣𝑖𝑑𝑡

= ቊ−2 𝑉 0 < 𝑡 < 2 𝑚𝑠+2 𝑉 2 < 𝑡 < 4 𝑚𝑠

𝑣𝑜 𝑡 =?


Differentiator Example

LCE.15

Solution (-30 mV)

𝑅 = 100 𝑘Ω, 𝐶 = 100 𝑛𝐹, 𝑣𝑖 𝑡 = 3𝑡 𝑉, 𝑣𝑜 𝑡 =?


Questions

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Dr. Gyurcsek István - vili.pmmf.huvili.pmmf.hu/~gyurcsek/PUB2GY/4.10X-LCE.pdf · Villamosságtan...

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