3. When a supply is added to a number of
resistors connected together end to end
the current can only take one
route through the circuit.
This is known as a series connection.
4. These rules apply to a series circuit:
The total circuit resistance (Rt), equals the
sum of all the circuit resistors. To work out
the total amount of circuit resistance, add
up all the individual resistor values.
Rt = R1 + R2 + R3 + etc……
5. The total circuit current (I), equals the
battery emf (V) divided by the total
resistance (R). Ie Ohm’s Law.
I = V
R
7. The potential difference across each
resistor is proportional to its resistance.
Voltage pushes the electrons through a
resistor. How much is used depends on
the size of the resistor. The bigger the
resistor the more we use.
So V1 = I x R1 , V2 = I x R2 etc……
8. The supply voltage will equal the sum of
all the potential differences across each
resistor.
If you add together all the potential
differences across each resistor (ie the
amount of volts dropped across each
resistor) it should equal the value of the
supply voltage.
Where V = V1 + V2 + V3 + etc….
9. The total power in a series circuit equals
the sum of all the individual powers used
by each resistor.
So P = P1+ P2 + P3 + etc…..
10. Example
A 5 ohm resistor is connected with
a 10 ohm resistor and a 12 V battery in
series.
5 Ω 10 Ω
12 V
11. (1) Find the total resistance.
To find Rt, add together all the values of
each resistance.
Rt = R1 + R2
Rt = 5Ω + 10 Ω
Rt = 15Ω
12. (2) Find the total current.
I = V
R
So I = 12
15
I = 0.8 Amps
13. (3)Find the potential difference across each
resistor.
Across R1
V = I x R
V1 = I x R1
V1 = 0.8 x 5
V1 = 4 Volts
Across R2
V = I x R
V2 = I x R2
V2 = 0.8 x 10
V2 = 8 Volts
14. (4) Find total voltage.
Add together all the voltage drops
Vt = V1 + V2
Vt = 4 + 8
Vt = 12V
15. (5) Find power used by each resistor.
Pt = V x I
Pt = 12 x 0.8
Pt = 9.6 Watts
16. (6) Find power used by R1
P1 = V1 x I
P1 = 4 x 0.8
P1 = 3.2 Watts
17. (7) Find power used by R2
P2 = V2 x I
P2 = 8 x 0.8
P2 = 6.4 Watts
