Copyright © 2013 The McGraw-Hill Companies, Inc...

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Transcript of Copyright © 2013 The McGraw-Hill Companies, Inc...

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  the ideal capacitor is a passive element with circuit symbol

  the current-voltage relation is

  the capacitance C is measured in farads (F) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2

�

i = C dvdt

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  capacitors can be bulky and typical values range from pF to µF

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Since then the energy stored in a capacitor is

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�

p(t) = i(t)v(t) = C dvdt

⎛ ⎝ ⎜

⎞ ⎠ ⎟ v =

dwdt

�

w = 12Cv

2

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  capacitors are open circuits to dc voltages   the voltage on a capacitor cannot jump

  capacitors store energy (iv>0) or deliver energy (iv<0)

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�

i = C dvdt

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Find i(t) for the voltages shown, if C=2 F.

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Solution: apply i(t)=2dv/dt and graph:

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Show that the following graphs are matching voltage and current graphs for a capacitor of C=5µF.

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Determine the maximum energy stored in the capacitor, and plot iR and iC.

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  the ideal inductor is a passive element with circuit symbol

  the current-voltage relation is

  the unit of inductance L is henry (H) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 10

�

v = L didt

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  inductors can be bulky and typical values range from µH to H

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Since then the energy stored in a inductor is

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�

p(t) = i(t)v(t) = L didt

⎛ ⎝ ⎜

⎞ ⎠ ⎟ i =

dwdt

�

w = 12 Li

2

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  inductors are short circuits to dc voltages   the current through an inductor cannot jump

  inductors store energy (iv>0) or deliver energy (iv<0)

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�

v = L didt

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Show that the following graphs are matching voltage and current graphs for an inductor of L=3 H.

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For the same 3-H inductor, the voltages are 10 times larger when the current is ramped 10 times faster:

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Determine the maximum energy stored in the inductor, and find the energy lost to resistor from t=0 to t=6 s.

Answer: 216 J, 43.2 J

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Apply KVL to show:

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�

Leq = L1 + L2 ++ LN

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Apply KCL to show

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�

Leq =1

1L1

+ 1L2

++ 1LN

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Apply KVL to show:

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�

Ceq =1

1C1

+ 1C2

++ 1CN

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Apply KCL to show:

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�

Ceq = C1 +C2 ++CN

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Two capacitors in series: Two inductors in parallel: Two resistors in parallel:

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�

Req =R1R2R1 + R2

�

Ceq =C1C2

C1 +C2

�

Leq =L1L2L1 + L2

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Show that these circuits are equivalent using series and parallel combinations.

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vout = !1

R1Cf

! t

0

vs dt "! vCf

(0)

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�

vout = −C1Rfdvsdt


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