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Electrotechnology
DC Circuit
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DC CircuitSpecific Learning Objectives
Apply circuit law to solve problem in dc network-series, parallel, combined dc and parallel circuits
Solve problem to find power and efficiency inelectromechanical system and consideration requiredwhen apply the laws in network problems
Solve problem regarding resistance required to extentrange of ammeter and voltmeter and verify the
experiments
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Electric Circuit.
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Electric Circuit
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Electric Circuit
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Electric Circuit
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Electric Circuit
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Electric Circuit
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Electric Circuit
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Electric Circuit
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DC Circuit
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DC Circuit
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DC CircuitOhms law
states that the current I flowing in a circuit is directly proportional to the appliedvoltage V and inversely proportional to the resistance R, provided thetemperature remains constant. Thus,
Electrical power
PowerPin an electrical circuit is given by the product of potential difference Vand current I. The unit of power is the watt, W. Hence
Electrical energy
Electrical energy = power time
If the power is measured in watts and the time in seconds then the unit of energy
is watt-seconds orjoules. If the power is measured in kilowatts and the time in hours then the unit of
energy is kilowatt-hours, often called the unit of electricity. The electricity meter in the home records the number of kilowatt-hours used
and is thus an energy meter.
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DC CircuitMain effects of electric current
The three main effects of an electric current are:
Magnetic effect : bells, relays, motors,
generators, transformers, telephones, car-ignition
and lifting magnets.
Chemical effect : primary and secondary cells and
electroplating
Heating effect : cookers, water heaters, electricfires, irons, furnaces, kettles and soldering irons
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DC Circuit Series circuits
Figure 5.1 shows three resistors R1,R2 and R3 connected end to end, i.e.,in series, with a battery source of Vvolts.
Since the circuit is closed a current I
will flow and the p.d. across eachresistor may be determined from thevoltmeter readings V1, V2 and V3
In a series circuit the current I is thesame in all parts of the circuit and
hence the same reading is found oneach of the two ammeters shown.
the sum of the voltages V1, V2 andV3 is equal to the total appliedvoltage
from Ohms Law :
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DC Circuit
Potential divider
The voltage distribution for thecircuit shown in Figure 2.2(a) isgiven by:Figure 2.2 : dividervoltage - series
Source : Electrical andElectronic 8th Edition (2002)
The circuit shown in Figure 2.2(a)is often referred to as a potentialdivider circuit. Such a circuit canconsist of a number of similar
elements in series connectedacross a voltage source, voltages
being taken from connectionsbetween the elements.
Frequently the dividerconsists of two resistors asshown in Figure 2.2(a), where
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DC CircuitParallel networks
Figure 2.3 shows three resistors, R1, R2and R3 connected across each other, i.e., in
parallel, across a battery source of V volts.
In a parallel circuit:
a) the sum of the currents I1, I2 and I3 isequal to the total circuit current, I, i.e.I=I1 +I2 +I3, and
b) the source p.d., V volts, is the sameacross each of the resistors.
From Ohms law:Figure 2.3 : Parallelcircuit
Source : Electrical and Electronic 8thEdition (2002)
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DC Circuit
Current division
For the circuit shown in Figure
2.4, the total circuit resistance,
RT is given by: Figure 2.4 :Parallel circuit
Source : Electrical and
Electronic 8th Edition (2002)
2.6, E1 is positive and E2 is
negative.)
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Light Circuit
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DC Circuit
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DC Circuit
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DC Circuit
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DC Circuit
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DC Circuit
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DC Circuit Kirchhoffs laws
Kirchhoffs laws state: Current Law. At any junction in
an electric circuit the total currentflowing towards that junction isequal to the total current flowing
away from the junction, Thus,referring to Figure 2.5:
or Voltage Law. In any closed loop
in a network, the algebraic sum ofthe voltage drops (i.e. products of
current and resistance) takenaround the loop is equal to theresultant e.m.f. acting in that loop.Thus, Referring to Figure 2.6:
Note that if current flows away from the positive terminal of a source,that source is considered by convention to be positive. Thus moving
anticlockwise around the loop of Figure
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DC Circuit
QUESTIONS
Define the Ohms Law and KirchoffsLaw.
Three resistors A = 50 , B = 20 & C= 25 . Calculate the total resistance if
the resistors connected in series andparallel.
A three resistor R1 = 2 , R2 = 3 andR3 = 8 are connected in series. Findthe voltage across each of the resistorsand supply voltage if the current supplyis 1.5 A.
For the network shown below I1 = 2.5A, I2 = -1.5 A. Calculate the current I3.
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DC Motor
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RESISTORS IN SERIES
A simple SERIES CIRCUIT is shown in the
diagram below. The current (I) at every point
in a series circuit equals the current leaving
the battery.
I1= I2=I3=ITotal
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RESISTORS IN SERIES
Assuming that the connecting wires offer no resistance tocurrent flow, the potential difference between the terminals
of the battery (V) equals the sum of the potential
differences across the resistors, i.e.,
V=Vl+ V2+ V3The equivalent electrical
resistance (R) for this
combination is equal to the
sum of the individualresistors, i.e.,
R=R1+ R2+ R3
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RESISTORS IN PARALLEL
In a simple PARALLEL CIRCUIT, the current leaving the
battery divides at junction point A in the diagram shown
below and recombines at point B. The battery current (I)
equals the sum of the currents in the branches. In general
I = I1 + I2 + I3
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RESISTORS IN PARALLEL
The potential
difference across each
resistor in thearrangement is the
same, i. e.
V = VI = V2 = V3
If no other resistance is present, the potential difference
across each resistor equals the potential difference across
the terminals of the battery. The equivalent resistance (R) of a parallel combination
is always less than the smallest of the individual
resistors. The formula for the equivalent resistance is as
follows: 1/R = 1/RI + 1/R2 + 1/R3
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RESISTORS IN PARALLEL
In a simple PARALLEL CIRCUIT, the current leaving the
battery divides at junction point A in the diagram shown
below and recombines at point B. The battery current (I)
equals the sum of the currents in the branches. In general
I = I1 + I2 + I3
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EMF AND TERMINAL
VOLTAGE All sources of emf have what is known as INTERNALRESISTANCE (r) to the flow of electric current. The internal
resistance of a fresh battery is usually small but increases
with use. Thus the voltage across the terminals of a battery is
less than the emf of the battery.
The TERMINALVOLTAGE (V) is given by the equation
V = - Ir, where represents the emf of the source of
potential in volts, I the current leaving the source of emf in
amperes and r the internal resistance in ohms.
The internal resistance of the source of emf is always
considered to be in a series with the external resistance
present in the electric circuit.
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KIRCHHOFF'S RULES
KIRCHHOFF'S RULES are used in conjunction with Ohm's
law in solving problems involving complex circuits:
KIRCHHOFF'S FIRST RULE or JUNCTION RULE: The
sum of all currents entering any junction point equals the sum
of all currents leaving the junction point. This rule is based on
the law of conservation of electric charge.
KIRCHHOFF'S SECOND RULE or LOOP RULE: The
algebraic sum of all the gains and losses of potential around
any closed path must equal zero. This law is based on the lawof conservation of energy.
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
1. Place a (+) sign next the long line of the battery symbol
and a (-) sign next to the short line. Start choosing a direction
for conventional current flow ( flow of positive charge )
If you choose the wrong direction for the flow of current in a
particular branch, your final answer for the current in that
branch will be negative. The negative answer indicates that
the current actually flows in the opposite direction.
I
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
2. Assign a direction to the circuit in each independent branch
of the circuit. Place a positive sign on the side of each resistor
where the current enters and a negative sign on the side where
the current exits, e.g.; This indicates that a drop in potential
occurs as the current passes through the resistor .
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
Notice how the
directions of the
currents are labeledin each branch of
the circuit
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
3. Select a JUNCTION POINT and apply the
junction rule, e.g., at point A in the diagram:
The junction rule may be applied at more than onejunction point. In general, apply the junction rule
to enough junctions so that each branch current
appears in at least one equation.
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
4. Apply Kirchhoffs loop rule by first taking notewhether there is a gain or loss of potential at each resistor
and source of emf as you trace the closed loop. Remember
that the sum of the gains and losses of potential must addto zero.
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
For example, for the left loop of
the sample circuit above start
at point B and travel
clockwise around the loop.
Because the direction chosen
for the loop is also the
direction assigned for the
current, there is a gain in
potential across the battery(- to +), but a loss of
potential across each resistor
(+ to -).
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
Following the path of the
current shown in the
diagram and using the
loop rule, the following
equation can be written:
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
The direction taken around the
loop is ARBITRARY. Tracing
a counterclockwise path around
the circuit starting at B, as
shown in the above right
diagram, there is gain in
potential across each resistor
to (- to +) and a drop in
potential across the battery (+to -). The loop equation would
then be:
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SUGGESTIONS FOR USING KIRCHHOFF'S
LAWS
Multiplying both sides of the above equation by - 1 and
algebraically rearranging, it can be shown that the two
equations are equivalent. Be sure to apply the loop rule to
enough closed loops so that each branch current appears in at
least one loop equation. Solve for each branch current using
standard algebraic methods.Solve
simultaneous
equations
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CAPACITORS IN SERIES AND
PARALLEL
A circuit with
CAPACITORS IN
PARALLEL is shown in thediagram below. According to
Kirchhoff s loop rule, the
potential difference (V) of the
source of emf:
V = VI = V2 = V3
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CAPACITORS IN PARALLEL
The total charge stored on the capacitor plates (Q) equals theamount of charge which left the source of:
Q = Ql + Q2 + Q3 ( Charge is additive)
and since Q = CV then CV = CV1 + CV2 + CV3
C= C1+ C2 +C3 (Capacitance is additive)
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CAPACITORS IN SERIES
For CAPACITORS IN SERIES,
the amount of charge (Q) that
leaves the source of emf equals
the amount of charge that forms
on each capacitor:
Q = Ql = Q2 = Q3
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CAPACITORS IN SERIES
From Kirchhoffs looprule, the potential
difference across the
source of emf (V) equals
the sum of the potentialdifferences across the
individual capacitors:
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Circuits containing resistors and
capacitorsAn RC CIRCUIT consists of a resistor and a capacitor
connected in series to a de power source.When switch
1 (S1), shown in the diagram below, is closed, thecurrent will begin to flow from the source of emf and
charge will begin to accumulate on the capacitor.
Using Kirchhoff s loop rule it can be shown that
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