Presentation is loading. Please wait.

Presentation is loading. Please wait.

Resistance in Series (1)

Similar presentations


Presentation on theme: "Resistance in Series (1)"— Presentation transcript:

1 Resistance in Series (1)
If we connect resistors across a source such that the ending point of one resistor is joined with starting point of the other resistor then they are said to be connected in series. The combined effect of all the resistors will be equal to the sum of individual resistances.

2 Consider two resistances R1 and R2
with terminals A, B and C, D as shown in the figure. A R1 B C R2 D

3 They will be in series if we connect
B with C as shown in the figure A R1 B C R2 D Combined effect of these two resistances will be Req = R1 +R2

4 Resistance in Series (2)
Therefore, if we connect N resistances in series then R eq = R1+R2+R3+…..+RN To illustrate this effect we take some examples.

5 Example 1 RAB→ So RAB= R1 +R2 =1 k+ 1k= 2k

6

7 Example 2 RAB 1 k  and 4 k  are in series so they
will be combined as

8 RAB→ Now 5k  and 3k  are also in series so

9 2 k  and 8k  are also in series
So RAB =2k + 8k =10k 

10 Consider two resistances with
terminals A, B and C, D as shown

11 If we connect A with C and B with
D they are said to be connected in parallel. The equivalent of these two resistances will be 1/Req = 1/R1 + 1/R2= (R1R2)/(R1+R2)

12 If we connect N number of resistances in parallel their equivalent will be
1/Req =1/R1+1/R2+1/R3+….+1/RN To illustrate this effect let us take some examples.

13 Example 3 1k is in parallel with 1k  so Req = (R1R2)/(R1+R2)=1/2 =0.5 k 

14

15 Example 4 4k  is parallel with 4k  so 4k||4k = (4 x 4)/(4 +4) =16/8= 2k 

16 2k  is parallel with 2k . So Req= (2 x2)/(2 +2)=4/4 =1 K 

17

18 12k  is parallel with 4k  so 12k||4k= (12x4)/(4 +12) =48/16=3k 
Example 5 12k 12k  is parallel with 4k  so 12k||4k= (12x4)/(4 +12) =48/16=3k 

19 Now 2k  is in series with 3k  so
RAB= 2k +3k=5k 

20

21 Example 6 4k  is in series with 8k  so the combined effect=12k 

22 12k  is in parallel with 12k  so 12k||12k=(12 x 12)/(12+12)

23 4k  is in series with 6k  so their combined effect =4k+6k =10k 

24 6k  is in parallel with 10k  so RAB = (6 x 10)/(6+10)=3.75k 

25 3.75k

26 Example 7 3k  is in series with 6k , therefore, their combined effect=3k +6k= 9k 

27 9k  is in parallel in 18k  so
9k||18k= (9 x 18)/(9+18) =162/27=6k 

28 6k  is in series with 10k  So their combined effect=6k+10k =16k 

29 6k  is in series with 16k  so
RAB= 6+16 =22k 

30

31 Example 8 1k  is in series with 2k  so Their combined effect= 1+2=3k 

32 3k  is in parallel with 6k 
3k||6k= (3 x 6)/(6+3) =18/9 =2k 

33 10k  is in series with 2k , therefore,
their combined effect =10k+2k =12k 

34 12k  is in parallel with 6k , hence
12k||6k= (12 x 6)/(12+6)=4k 

35 2k  is in series with 4k  combined effect= 2+4=6k 

36 6k  is in parallel with 6k , therefore,
6k||6k=(6 x 6)/(6+6)=36/12=3k 

37 3k  is in series with 9k , therefore,
combined effect=3k +9k=12k 

38 12k  is in parallel with 4k  so
12k||4k = (12 x 4)/(12+4)=48/16=3k 

39 2k  is in series with 3k  so
RAB=2k+3k=5k 

40


Download ppt "Resistance in Series (1)"

Similar presentations


Ads by Google