6Ω
12Ω
6Ω
Resistors in series cause more resistance to the current.
Combined resistance equals
the sum of all resistances.
RT = R1 + R2 + R3 + … Rn
RT = 2 + 3 + 4 = 9Ω
Combined resistance also called:
Effective resistance
Total resistance
1) Find the combined resistance of the following circuit.
RT = R1 + R 2 + R 3 =
10 + 5 + 9 = 24 Ω
R1
R2
Important to name
the resistors
R3
2) Find the currents in the ammeters if the current in R3 is 2A.
Series circuit: I1 = I2 = I3 = 2A
3) Find the voltage on R1 , R2, R3
I1 = 2 A
R1 = 5 Ω
R2 = 10 Ω
R3 = 9 Ω
V1 = ?
V2 = ?
V3 = ?
V1 = I1 x R1 = 2 x 5 = 10V
V2 = I2 x R2 = 2 x 10 = 20V
V3 = I3 x R3 = 2 x 9 = 18V
From the following circuit, find:
a) the combined resistance
b) the current flowing through R1 and R2
c) the voltage across R1
d) the voltage across R2
4v
R1 = 6 Ω
a) RT = R1 + R2 = 6 + 2 = 8 Ω
R2 = 2 Ω
b) IT = VT / RT = 4 / 8 = 0.5 A
Series circuit: IT = I1 = I2 = 0.5 A
c) V1 = I1 x R1 = 0.5 x 6 = 3V
d) V2 = I2 x R2 = 0.5 x 2 = 1V
Check yourself:
VT = V1 + V2
?
4 = 3 + 1

a) RT = R1 + R2 + R3 = 10 + 6 + 8 = 24 Ω
b) IT = VT / RT = 24 / 24 = 1 A
Series circuit: IT = I1 = I2 = I3 = 1 A
c) V1 = I1 x R1 = 1 x 10 = 10V
V2 = I2 x R2 = 1 x 6 = 6V
V3 = I3 x R3 = 1 x 8 = 8V
Check yourself:
VT = V1+V2+V3 = ?
24 = 10+6+8

Good way to
find mistakes
The combined resistance is smaller than the smallest resistor.
4v
Combined resistance:
1
1
1
1
1



 ...
RT R1 R2 R3
Rn
R1 = 6 Ω
R2 = 2 Ω
1
1
1 1 1 26 8 2
 
  


RT R1 R2 6 2 2  6 12 3
3
RT   1.5
2
Smaller than the
smallest resistor:
1.5 < 2

1) a) Find the combined resistance of the following circuit.
20 v
1
1
1
1
1




30 60
RT
R1 R2
90
1
30  60


1800
RT
30 x 60
RT 
1800
 20
90
= 30Ω
Smaller
than the
smallest
(30Ω)
b) Find the voltage across V1 and V2
Parallel circuits:

= 60Ω
V1 = V2 = VT = 20v
c) Find the currents I1 , I2 and IT.
V
20 2
V
20 1
I1 

 A
I2 

 A
R1 30 3
R2 60 3
IT = I1 + I2 = ⅔ + ⅓ = 1A
20 v
R1  R2
RT 
R1  R2
= 30Ω
= 60Ω
1
1
1
R1  R2



RT
R1 R2
R1  R2
R1  R2
RT 
R1  R2
Find the effective resistance in the following
circuit
R1  R2
RT 
R1  R2
500  1200 600000
RT 

5001  1200
1700
RT  352.941
Smaller than
the smallest
(500Ω)

A real circuit is usually mix of series and parallel in one circuit.
1) Divide the circuits to small parts.
2) Start with parts that do not affect other parts.
3) Simplify the circuits as you go.
Find the combined resistance
RB 
RA = R2 + R3 = 8 Ω + 4 Ω = 12 Ω
RT = Rcomb + R1 = 6 Ω + 6 Ω = 12 Ω
RA xR4
12 x12

 6
RA  R4 12  12
60 v
1) From the following circuit, find:
A
(Assume that the switch is closed)
A. the combined resistance
B. What is the reading in the ammeter?
C. the voltage across R1, R2, R3, R4
D. the current in R1, R2, R3, R4
E. What will be the reading in the ammeter,
if the switch is open?
Show your working!
R1=10Ω
R3=20Ω
R2=20Ω
R4=40Ω
60 v
1) A. Find the combined resistance
R1,2 = (10x20) / (10+20) = 20/3Ω
A
R1=10Ω
R3=20Ω
R2=20Ω
R4=40Ω
R3,4 = (20x40) / (20+40) = 40/3Ω
RT = R1,2 + R3,4 = 20/3 + 40/3 = 60/3 = 20Ω
B. What is the reading in the ammeter?
IT = VT / RT = 60/20 = 3A
60 v
C. Find the voltage across R1, R2, R3, R4
A
V1 = V2 = V1,2 = I x R1,2 = 3 x 20/3 = 20V
R1=10Ω
R3=20Ω
R2=20Ω
R4=40Ω
V3 = V4 = V3,4 = I x R3,4 = 3 x 40/3 = 40V
(20 + 40 = 60 V  that means we are correct)
D. Find the current in R1, R2, R3, R4
I1 = V1 / R1 = 20/10 = 2A
I2 = V2 / R2 = 20/20 = 1A
I3 = V3 / R3 = 40/20 = 2A
I4 = V4 / R4 = 40/40 = 1A
E. What will be the reading in the
ammeter, if the switch is open?
IT = 0A  The circuit is open,
no current passes through it.
2) The following diagram shows 4
identical bulbs connected to a
battery.
Which bulbs will continue to work if:
L1
A. Bulb L2 will be removed from the circuit?
L1 and L4 continue to work
B. Bulb L4 will be removed from the circuit?
L1 , L2, L3 continue to work
C. Bulb L1 will be removed from the circuit?
All bulbs turn off
L2
L3
L4
3) From the following circuit, calculate the combined resistance
between points A and B when:
a) S1, S2 and S3 are closed
b) S2, S3 are closed while S1 is open
c) S1, S3 are closed while S2 is open
A. RT = R2||R3
S1
A
R2 = 3Ω
R1 = 6Ω
R3 = 6Ω
S2
S3
*(S1 cause a short circuit over R1 so no current passes through R1)
RT = (3x6)/(3+6) = 2Ω
B. RT = R1 + R2||R3
*(Current passes through all resistors)
RT = 6 + (3x6)/(3+6) = 8Ω
C. RT = R3 +R1
*(Current passes through R3 AND R1)
RT = 12Ω
B
http://www.teacherrambo.com/_MyStuff/M3U1Resistors-Storyline/story_html5.html
Circuits
Series
Parallel
IT = I1 = I2 = I3= In
VT = V1 + V2 + V3 + Vn
IT = I1 + I2 + I3+ In
VT = V1 = V2 = V3 = Vn
RT = R1 + R2 + R3 +…Rn
1
1 1 1
1
    ...
RT R1 R2 R3
Rn
2
resistors
ONLY
RT 
R1  R2
R1  R2