Page 37 - Schaum's Outline of Theory and Problems of Electric Circuits

CIRCUIT LAWS
26
v ¼ iR 1 þ iR 2 þ iR 3 [CHAP. 3
¼ iðR þ R þ R Þ
3
2
1
¼ iR eq
where a single equivalent resistance R eq replaces the three series resistors. The same relationship
between i and v will pertain.
For any number of resistors in series, we have R eq ¼ R þ R þ .
2
1
If the three passive elements are inductances,
di di di
v ¼ L 1 þ L 2 þ L 3
dt dt dt
di
¼ðL 1 þ L 2 þ L 3 Þ
dt
di
¼ L eq
dt
Extending this to any number of inductances in series, we have L eq ¼ L 1 þ L 2 þ .
If the three circuit elements are capacitances, assuming zero initial charges so that the constants of
integration are zero,
ð ð ð
1 1 1
v ¼ idt þ idt þ idt
C 1 C 2 C 3
ð
1 1 1
¼ þ þ idt
C 1 C 2 C 3
ð
1
¼ idt
C eq
The equivalent capacitance of several capacitances in series is 1=C eq ¼ 1=C 1 þ 1=C 2 þ .

EXAMPLE 3.3. The equivalent resistance of three resistors in series is 750.0
. Two of the resistors are 40.0 and
410.0
. What must be the ohmic resistance of the third resistor?

R eq ¼ R 1 þ R 2 þ R 3
and R 3 ¼ 300:0
750:0 ¼ 40:0 þ 410:0 þ R 3

EXAMPLE 3.4. Two capacitors, C 1 ¼ 2:0 mF and C 2 ¼ 10:0 mF, are connected in series. Find the equivalent
capacitance. Repeat if C 2 is 10.0 pF.
6
6
ð2:0 10 Þð10:0 10 Þ
C 1 C 2
C eq ¼ ¼ 6 6 ¼ 1:67 mF
C 1 þ C 2 2:0 10 þ 10:0 10
If C 2 ¼ 10:0 pF,
6
ð2:0 10 Þð10:0 10 12 Þ 20:0 10 18
C eq ¼ ¼ ¼ 10:0pF
2:0 10 6 þ 10:0 10 12 2:0 10 6
12
where the contribution of 10:0 10 to the sum C 1 þ C 2 in the denominator is negligible and therefore it can be
omitted.
Note: When two capacitors in series differ by a large amount, the equivalent capacitance is essen-
tially equal to the value of the smaller of the two.

3.5 CIRCUIT ELEMENTS IN PARALLEL

For three circuit elements connected in parallel as shown in Fig. 3-4, KCL states that the current i
entering the principal node is the sum of the three currents leaving the node through the branches.