The document introduces series and parallel circuits. It explains that for series circuits, total resistance is calculated by adding individual resistances. For parallel circuits, total resistance is the reciprocal of the sum of the reciprocals of individual resistances. Ohm's law relates current, voltage and resistance. Examples are provided to calculate total resistance, current and power in circuits with both series and parallel components using Kirchhoff's laws.

1. Introduction to series and parallel circuitry.

2. Calculations in Series & Parallel
Resistor Networks
Resistors in series:
The total resistance of two or more resistors connected in series is calculated by simply adding the individual
values of resistances to find the sum total (RTotal):
Total Resistance for any number of resistors in series = R1 + R2 + R3 …
Example
Adding resistors in series always
increases the total equivalent
resistance.
Self check:
Your total resistance result should
always be larger then the largest
resistor in the circuit.
25Ω > 15 Ω 
Can you think why this is so?

3. Resistors in parallel:
To calculate the total resistance of a circuit of parallel resistors the following steps can be used:
-Firstly calculate the sum of the reciprocals of each resistors value:
𝑠𝑢𝑚 =
1
𝑅1
+
1
𝑅2
+
1
𝑅3
⋯
-Secondly take the reciprocal of this sum to find RTotal :
𝑅Total =
1
𝑠𝑢𝑚
Example
Adding resistors in parallel always
decreases the total equivalent
resistance.
Self check:
Your total resistance result should
always be smaller then the smallest
resistor in the circuit.
12.8Ω < 22 Ω 
Can you think why this is so?

4. Introduction to Ohm’s Law
• Ohms law states that the current (I) through a conductor is directly
proportional to the voltage (V) applied across the conductor, and
inversely proportional to the conductor’s resistance (R).
• Three equivalent expressions of Ohm's law are:
𝐈 =
𝐕
𝐑
𝐕 = 𝐈 × 𝐑 𝐑 =
𝐕
𝐈
𝑉
V
I R×

5. Using Ohm’s Law
Example 1 – Calculating a circuit’s loop current.
V = 12 V R = 8 Ω
I = ?
𝑉
V
I R 𝑉
12
? 8
I =
12 V
8 Ω
𝐈 =
𝐕
𝐑
= 1.5 A

6. Series and Parallel resistor circuit
Calculate the Total Current for the network
6
R1
R5
R3
6V
R2
R6
R4
94Ω
94kΩ
64kΩ
61Ω
82kΩ
68Ω
𝑉
V
I R

7. R1
R5
R3
6V
94Ω
94Ω
64Ω
Series Parallel
R2
R4
R6
R4
R2
R6
Ra=R2+R4+R6
61Ω
82Ω
68Ω
=211Ω
Series and Parallel resistor circuit
Calculate the Total resistance for the network

8. = R1 + ( 𝐼𝐼 ) + R5
1
𝑅𝑏
= 94 + Rb + 94
= 237.11Ω
Calculate the Total resistance for the network
𝑅6
𝑅4+𝑅2+𝑅3
R1
R5
R3
6V
94Ω
94Ω
64Ω
R2
R4
R6
Ra211Ω
Ra=R2+R4+R6 = 211Ω
=
1
𝑅3
+
1
𝑅 𝑎
1
𝑅𝑏
=
1
64
+
1
211
= 49.1054Ω
= 94 + 49.11 + 94

9. Calculate the Total current for the network
𝑉
V
I R
IT =
6
237.11
IT =
Vs
RT
= 0.0253A = 25.31mA
R1
R5
R3
6V
94Ω
94Ω
64Ω
R2
R4
R6
Ra211Ω
= 237.11Ω= 94 + 49.11 + 94

10. Calculate the current Flowing through R3
1
Rb
=
1
64
+
1
211
= 49.1054Ω
R1
R5
R3
6V
94Ω
94Ω
64Ω
R2
R4
R6
Ra211Ω
𝑉
V
I R
V= 𝐈 𝐱 𝐑
Vdb= 0.0253 It 𝐱 49.1054(𝑅𝑏)= 1.24V
I =
1.24
64
IR3 =
V3
R3
= 0.0194A = 19.42mA
(R3)
It =
6
237.11
= 0.0253A = 25.31mA
Rb
(Vdb)

11. Calculate Potential Difference across R1 , R5 and
𝑉
V
I R
V𝐝𝟏 = 𝐈𝐭 𝐱 𝐑𝟏
R1
R5
R3
6V
94Ω
94Ω
64Ω
R2
R4
R6
Ra211Ω
V𝐝𝐛 = 0.0253 It 𝐱 49.1054(𝑅𝑏)= 1.24V
Vd1= 0.0253 It 𝐱 94(𝑅1)= 2.38V
V𝐝𝟓 = 0.0253 It 𝐱 94(𝑅5)= 2.38V
Check
Adding all the potential differences should equal
to the supply voltage
Vs=Vd1+Vd5+VRb
Vs= 2.38 + 2.38 + 1.243
VS=6V

12. Calculate the Current flowing throw R3
R1
R5
R3
6Vs
94Ω
94Ω
64Ω
R2
R4
R6
Ra211Ω
IR3 =
Vdb
R3
=
1.24
64
= 0.0194A = 19.4mA
V𝐝𝐛 = 0.0253 It 𝐱 49.1054(𝑅𝑏)= 1.24V

13. Calculate Total power dissipated by the network
Power = Current (IT) x Voltage (Vs = Watts (W)
W= 0.0253 𝐱 6 = 0.151.86 = 151.86mW

14. Calculate power dissipated by R3
PR3 =
(V)2
R3
=
1.244 2
64
= 0.02417 w
= 24.17mw
Power = Current x Voltage = Watts (W)
W= 0. 01942 𝐱 1.24 = 24.081mW
OR

15. http://tinyurl.com/hz7g7fv
Check results using Falstad

16. Kirchhoff’s Current Law (KCL)
Given the following values, calculate the current flowing
through R3.
V1 = 21 V, V2 = 11 V, R1 = 4 Ω, R2 = 14 Ω and R3 = 25 Ω.
The circuit below contains 2 power sources.
Kirchhoff’s Current Law:- “The algebraic sum of all currents entering and exiting a node must equal zero”.
Ientering + (-Iexiting )= 0

17. Kirchhoff’s Voltage Law (KVL)
21v 11v
4 Ω
25 Ω
14 Ω𝑰 𝟏 𝑰 𝟐
𝑰 𝟑
𝐼3= 𝐼1 + 𝐼2
Loop 1 Loop 2
Given the following values, calculate the current flowing through R3.
Loop 1 V1 = I1R1 + I3R3
V1 = I1R1 + (I1 + I2) R3
V1 = I1R1 + I1R3 + I2R3
21 = I14 + I125 + I225
21 = I14 + I125 + I225
21 = I129 + I225
Using Kirchhoff's Voltage Law, VCL the equations are given as:- V1 = VR1 + VR3
Loop 2 V2 = I2R2 + I3R3
V2 = I2R2 + (I1 + I2) R3
V2 = I2R2 + I1R3 + I2R3
11 = I214 + I125 + I225
11 = I214 + I125 + I225
11 = I239 + I125
As I3 = I1 + I2 we can rewrite the equations as
Add together similar factors in the equations
Substitute values for Volts and Resistance
V1 = 21 V, V2 = 11 V, R1 = 4 Ω, R2 = 14 Ω and R3 = 25 Ω.
𝑉
V
I R

18. Kirchhoff’s Voltage Law (KVL)
Loop 1 equation 21 = I129 + I225 Loop 2 equation 11 = I239 + I125
Now we have two “Simultaneous Equations” that can be reduced to give us the values of I1 and I2
If equation L1 is multiplied throughout by (25) and equation L2 by (29)
Loop 1 21 = I129 + I225 (x 25)
Loop 2 11 = I239 + I125 (x 29)
Then the coefficient of I1 will be the same in the newly formed equations
Loop 1 525 = I1725 + I2625
Loop 2 319 = I21131 + I1725
Subtracting equation L1 from equation L2
206 = I2 -506
Transpose this formula to obtain one for I2 .
206 = I2
-506
Rearrange
I2 = 206 I2 = -0.407A
-506

19. Kirchhoff’s Current Law (KCL)
Calculate the current flowing through R3.
Substitution of I2 to give the value of I3 giving the value I2 = -0.407A
Loop 1 21 = I129 + I225
21 = I129 + (-0.407x25)
21 = I129 + (-10.175)
10.175 + 21 = I129
31.175 = I129
I1 = 31.175 I1 = 1.075A
29
I3 = I1 + I2
I1 = 1.075A, I2 = -0.407A
I3 = 1.075 + (-0.407)
I3 = 0.688A

20. Kirchhoff's Voltage Law, KVL
The current flowing in R3 resistor is given as :-
1.075 + (-0.407) = 0.688A
Resistance R3 = 25Ω
Power distributed across R3
V = IR
V = 0.668 x 25
V = 16.7V
Power = Current x Voltage = Watts (W)
W = 0.688 x 16.7
W = 11.15W

21. Check results using Falstad
http://tinyurl.com/ly5z4jb