electrical circuitsand methods of network analysis

Electrical CurrentNetwork analysis

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

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


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


Reduce and Return Approach

• For


Reduce and Return Approach• Ex. For circuit in Fig. find branch currents.

• CDR

• KCL


Reduce and Return Approach• Ex. Calculate the total resistance, current supplied by the source, branch currents

and the voltage across R6.

•

•


Reduce and Return Approach•

• Working back to I6

•

•

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


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

•


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


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

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

Loop Current Analysis Method

apply loop current method to the network

Loop Current Analysis Method.

Ex

I1 = 1 A and I2 = 2 A

Electrical Circuits

DC (direct current): If the current flowing through an element is constant and of unidirectional

-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

Electrical CurrentElectric 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

Direction of Electrical Current

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

VoltageBasic 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.

Polarity of VoltagePotential: 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

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

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.

Plotting Ohm’s law

Power and Energy

Energy is the capacity to perform work.

Power is the time rate of expending or absorbing energy.

Power and Energy

Ex: Find the power delivered to the dc motor of Fig. 1.5

Energy ConversionTo 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 .

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.

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 =

EfficiencyThe conservation of energy requires that

Energy input = Energy output + Energy lost or stored in the system

Win = Wout + Wlost or stored by the system

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?

electrical circuitsand methods of network analysis

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