13.10.2012 Views

boylistad

boylistad

boylistad

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

410 ⏐⏐⏐ CAPACITORS<br />

+<br />

E<br />

–<br />

Q T<br />

V 1<br />

Q 1<br />

V 2<br />

Q 2<br />

+ – + – + –<br />

FIG. 10.61<br />

Series capacitors.<br />

V 3<br />

Q 3<br />

time between 0 and t 1. The same can be said about any portion of the<br />

voltage waveform that has a constant slope.<br />

An important point to be gained from this discussion is that it is not<br />

the magnitude of the voltage across a capacitor that determines the current<br />

but rather how quickly the voltage changes across the capacitor. An<br />

applied steady dc voltage of 10,000 V would (ideally) not create any<br />

flow of charge (current), but a change in voltage of 1 V in a very brief<br />

period of time could create a significant current.<br />

The method described above is only for waveforms with straight-line<br />

(linear) segments. For nonlinear (curved) waveforms, a method of calculus<br />

(differentiation) must be employed.<br />

10.13 CAPACITORS IN SERIES AND PARALLEL<br />

Capacitors, like resistors, can be placed in series and in parallel.<br />

Increasing levels of capacitance can be obtained by placing capacitors<br />

in parallel, while decreasing levels can be obtained by placing capacitors<br />

in series.<br />

For capacitors in series, the charge is the same on each capacitor<br />

(Fig. 10.61):<br />

Applying Kirchhoff’s voltage law around the closed loop gives<br />

E � V1 � V2 � V3 However, V � � Q<br />

C �<br />

so that<br />

QT �<br />

CT<br />

Q1 Q2 Q3 ������ C1 C2 C3<br />

Using Eq. (10.29) and dividing both sides by Q yields<br />

(10.29)<br />

(10.30)<br />

which is similar to the manner in which we found the total resistance of<br />

a parallel resistive circuit. The total capacitance of two capacitors in<br />

series is<br />

(10.31)<br />

The voltage across each capacitor of Fig. 10.61 can be found by first<br />

recognizing that<br />

QT � Q1 or C T E � C 1V 1<br />

Solving for V 1:<br />

Q T � Q 1 � Q 2 � Q 3<br />

1 1 1 1<br />

�� � �� � �� � �� C C C C<br />

T<br />

1<br />

2<br />

C<br />

CT � �<br />

1C2 �<br />

C1 �<br />

V1 � CTE �C1<br />

C 2<br />

3

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!