An electric circuit is a path in which electrons from a voltage or current source flow. The point where those electrons enter an electrical circuit is called the "source" of electrons.

1. Electrical Current
Network analysis

2. Methods of network analysis
SERIES-PARALLEL NETWORKS
• Networks that contain both series and parallel circuit
configurations
• A firm understanding of the basic principles associated
with series and parallel circuits is a sufficient background
to begin
– Investigation -analysis - and design of networks
• One can become proficient in the analysis of series-parallel
networks only through exposure, practice, and experience
• There are a few steps that can be helpful in getting started for
network analysis

3. General approach for network analysis
General approach
• Try to visualize the network problem “in total”
• make a brief mental sketch of the overall approach you plan to
use.
• Next examine each region of the network independently before
tying them together in series-parallel combinations. This will
usually simplify the network and possibly reveal a direct
approach toward obtaining one or more desired unknowns.
• Redraw the network with the reduced branches.
• When you have a solution, check that it is reasonable

4. General approach for network analysis
Reduce and Return Approach
• For many single-source, series-parallel networks, the analysis is one
that works back to the source, determines the source current, and
then finds its way to the desired unknown
• First, series and parallel elements must be combined to establish the
reduced circuit
• source current can now be determined using Ohm’s law
• Then one can proceed back through the network
• When you have a solution, check that it is reasonable

5. General approach for network analysis
Reduce and Return Approach
• For

6. General approach for network analysis
Reduce and Return Approach
• Ex. For circuit in Fig. find branch currents.
• CDR
• KCL

7. General approach for network analysis
Reduce and Return Approach
• Ex. Calculate the total resistance, current supplied by the source, branch
currents and the voltage across R6.
•
•

8. General approach for network analysis
Reduce and Return Approach
•
• Working back to I6
•
•

9. Branch Current Analysis Method
• will produce branch currents
• then other quantities, such as voltage, power, etc can be determined
Steps
• Assign a distinct current of arbitrary direction in each branch of the network.
• Indicate the polarities of voltage drops for each resistor as determined by
the assumed current direction.
• Apply KVL around each closed, independent loop of the network.
• Apply KCL at the minimum number of nodes that will include all branches
• Solve the simultaneous linear equations for assumed branch currents

10. Branch Current Analysis Method
Apply the branch current method
• Step 1: Three branches cda, cba and ca
• Three currents of arbitrary direction I1, I2, and I3 are chosen.
• Current directions are chosen to match voltage sources E1 and E2
• Step 2: Polarities of voltage drops for each resistor are drawn
• Step 3: Applying KVL around each loop in clockwise direction
• Apply KVL
•

11. Branch Current Analysis Method
Apply KCL
• Step 5: Three equations and three unknown. So, the parameters can
be obtained by solving the three equations.
• Therefore, I1 = - 1A I2 = 2 A I3 = - 1 A

12. Branch Current Analysis Method
Ex. Find branch currents applying branch current method to network
KVL results
Solving
I1 = 4.773 A
I2 = 7.182 A
I3 = 2.409 A

13. Loop Current Analysis Method
• suitable for coupled circuit solutions
• employs a system of loop or mesh currents instead of branch
currents
• currents in different meshes are assigned continuous paths so that
they do not split at a junction into branch currents
• best suited when energy sources are voltage sources
• consists of writing loop voltage equations by KVL in terms of loop
currents
Steps
Step 1: Assign a distinct current in the clockwise direction to each
independent, closed loop of the network.
Step 2: Indicate the polarities within each loop for each resistor as
determined by the assumed direction of loop current for that loop.
Step 3: Apply Kirchhoff’s voltage law around each closed loop in the
clockwise direction.
Step 4: Solve the resulting simultaneous linear equations for the
assumed loop currents

14. Loop Current Analysis Method
apply loop current method to the network

15. Loop Current Analysis Method
.
Ex
I1
= 1 A and I2
= 2 A

17. Electrical Circuits
DC (direct current): If the current flowing
through an element is constant and of
unidirectional

18. Electrical Circuits
DC (direct current): If the current flowing
through an element is constant and of
unidirectional

19. -Electric Current results from charges in motion
- Flow of current is flow of positive charges.
- Charge is the intrinsic property of matter and expressed in terms of charge of one
electron, e= -1.602X10-19 C
- -1C = charge on 6.24X1018 electrons
-Description of current requires a value and a direction (indicated by
arrow)
Current flows through a specified area and is defined by the
electric charge passing through the area per unit time
Unit of current is Ampere (A). 1A = 1 C/s
1 A = 1 C charge moving across a fixed surface in 1 s
Electric Current

20. Electrical Current
Electric Current: 2 types
DC (direct current): If the current flowing through an
element is constant and of unidirectional
AC (alternating current): A time varying current i(t) :
Such as a ramp, a sinusoidal or an exponential

21. Direction of Electrical Current
Electric Current always flows from + (positive)
terminal of battery to – (negative) terminal through
external circuit

22. Voltage
Basic variables of an electrical circuit: Current and Voltage.
Voltage across an element is the work (energy) requires to move
a unit positive charge from – (negative) terminal to + (positive)
terminal.
• Unit of voltage is volt (V).
• If 1 J of work is required to move the 1 C charge from one
position, let A to another B, then position B is at a potential of 1 V
with respect to position A.
v = dw/dq v = voltage,
w = work or energy and
q = charge.

23. Polarity of Voltage
Potential: The voltage at a point with respect to another point.
Potential difference: The algebraic difference in potential (or voltage)
between two points of a network.
Voltage: When isolated, like potential, the voltage at a point with respect
to some reference such as ground (0 V).
Voltage difference: The algebraic difference in voltage (or potential)
between two points of the system (drop or rise).
• Voltage source (i.e. battery) pressures or established current from – (negative)
to + (positive) terminal of the battery
• Polarity of the voltage drop across the resistor: Current enters into + terminal of
an element and exits from - terminal

24. Ohm’s law
Ohm’s law relates between the current flow through a conductor and the
voltage applied across the conductor.
Ohm’s law states that
The ratio of the potential difference (E) between any two
points on a conductor to the current (I) flowing between
them, is constant, provided the temperature of the conductor
does not change.
Magic Triangle Simple Circuit

25. Ohm’s law
Ex: Determine the current resulting from the application of a 9 V battery across a
network with a resistance of 2.2 kΩ.

26. Plotting Ohm’s law

27. Power and Energy
Energy is the capacity to perform work.
Power is the time rate of expending or absorbing energy.

28. Power and Energy
Ex: Find the power delivered to the dc motor of Fig. 1.5

29. Energy Conversion
To produce an energy conversion (heat, light, motion, etc), power must be used over
a period of time. A motor may have the horsepower (HP) to run a heavy load, but
unless the motor is used over a period of time, there will be no energy conversion. the
longer the motor is used to drive the load, the greater will be the energy expended.
Power is the rate of work done.
Instantaneous power, p = dW/dt = (dW/dq) (dq/dt) = v.i
So . dw = p dt
By integrating, Total work done, W = ∫ p dt .

30. Energy Conversion
Energy (W) lost or gained by any system
Ws is too small a quantity Watthour (Wh) and kilowatthour (kWh) are used
Ws or J
Ex: What is the cost of using a 5 HP motor for 2 h if the rate is Tk. 5 per kWh?
Cost = (7.46 kWh)(5 Tk/kWh) = 37.30 Tk.

31. Energy Conversion
Energy (W) lost or gained
Practical unit used by Power sectors Wh and kWh
Ws or J
Ex: What is the total cost of using all of the following at Tk 5 per kWh.
1200 W toaster for 30 min, six 50 W bulbs for 4 h, a 400 W washing machine for 45 min,
and a 4800 W electric clothes dryer for 20 min.
so, Cost = (3.7 kWh) ( 5 Tk/kWh) = 18.50 Tk
W =

32. Efficiency
The conservation of energy requires that
Energy input = Energy output + Energy lost or stored in the system
Win = Wout + Wlost or stored by the system

33. Efficiency
Ex: A 2-hp motor operates at an efficiency of 75%. What is the power input in
watts? If the applied voltage is 220 V, what is the input current?