W9_parallel_resonance.eps

178 Week 5: Resistance

I

R

R

I R I R I R

b

I

R

b

a

a

I

R R

a b

a b

I

tot

tot

1 2 3

tot

tot

tot

tot

1 2 3

1 2 3

(a) (b)

Figure 56: Three resistorsR 1 , R 2 , R 3 arranged inseries(left, (a)) andparallel(right, (b)), along
with the equivalent/total resistances of each one portrayed below. In both cases the total resistance
is “equivalent” when applying a voltageVabacross theaandbcontacts produces thesame total
currentItotin the top and bottom figure.

current:

V 1 = ItotR 1 (327)

V 2 = ItotR 2 (328)

V 3 = ItotR 3 (329)

Obviously the total voltageVabis given by:

Vab=V 1 +V 2 +V 3 =Itot(R 1 +R 2 +R 3 ) (330)

If we look at the lower (a) diagram, Ohm’s Law yields:

Vab=ItotRtot (331)

Equating and cancelling the commonItot, we get:

Rtot=R 1 +R 2 +R 3 (332)

There was nothing “special” about having only three resistors. We could have had, four, five, or
Nresistors in series and we’d simply have more terms in a general equation:

Vab=

∑N

i=1

ItotRi=Itot

∑N

i=1

Ri=ItotRtot (333)

so thatin generalthe rule for the addition ofNresistors in series is:

Rtot=R 1 +R 2 +...+RN=

∑N

i=1

Ri (334)

Parallel

In the case of resistances in parallel, we have thesamevoltageVabapplied across all of the resistors
in parallel. If we look at the upper (b) figure, we can use Ohm’s Law to evaluate thecurrentthrough