RESISTANCE AND CIRCUIT ANALYSIS

EQUIVALENT RESISTANCE

| A single resistance that can replace all the resistances in an electrical circuit while maintaining the same current when connected to the same source

| Symbol is Rtotal or Rt

RESISTORS IN CIRCUITS

| Resistors in Series y substitute Ohm’s Law into KVL y generates an equivalent resistance y Rt = Rseries = R1 + R2 + R3 + .......

RESISTORS IN CIRCUITS

| Resistors in parallel y Substitute Ohm’s Law (isolate I) into KCL y generate an equivalent resistance y Equation:

1 2 3

1 1 1 1 1 ...t parallelR R R R R � � �

EXAMPLE #2

| Determine the equivalent resistance for a 25.2 ohm resistor connected in series with a 28.12 ohm resistor. R ,

=

25*2:2=28 .br

Regis RitR.:ReE53Br= 25.2+28,12

= 53.32

EXAMPLE #3

| Determine the equivalent resistance of a 120 ohm resistor connected in parallel with a 60 ohm resistor. ieikiie

Itsstir intothaottoka =EoReq ÷he? Hk

EQUIVALENT RESISTANCE IN MIXED CIRCUITS

Sn +

Sr aRp=¥ti5 Ss

sr±st 's

we ¥,.£

Sr Sr Rp = } 5r5r Sr Ss

Rp=2 .sr

Req=RitRptR4tRs b= 5+2.5+5+5 pptetsttsttstts

Req=h .sr REGIRITRP

=5tt2sk=±sReq=6.25k Rp=EReqi 6.2g Rp=l.Kr

( zsigdigf

EQUIVALENT RESISTANCE IN MIXED CIRCUITS

CIRCUIT ANALYSIS

| To analyse a circuit means to find all the unknown values of voltage, current, and resistance.

| Use a combination of equivalent resistance, Kirchoff’s laws, and Ohm’s Law.

CIRCUIT ANALYSIS EXAMPLE

VFR

V

IR

±s=V=6-

12 .0r Source

Gora"Lb18r

=l{ a

0.33A

" " 6 V¥¥§d 1

kovtkaalz.or assn .

A WE v,=xN 2 2.0Ybnkor03.3 12 .or 3,Y3Zortuakor. 0.33A ftEE Iz=V+6 .0✓ 12 .0r =D Rr

4.0 ZV =¥a

'

Ttejtzttz YeReq=RitRp +=2# '÷¥

=12rt6r Rp =Ya=18r Rp=6R =yba

WORK

| Read 7.4 and 7.6 | Pg. 337 # 1,2 | Pg. 339 # 4,6 | Pg. 340 #7 | Pg. 342 #9

| Handout Assignment Due Tuesday | Work day tomorrow with lots of opportunity for

practice !!!!