KEMAHIRAN MANIPULATIF I

1
KEMAHIRAN MANIPULATIF I
EDU3416
The amount of current I in a circuit depends on
its resistance R and the applied voltage. I
V/R If we keep the same resistance in a circuit
but vary the voltage, the current will vary.
THINK! 1000v applied across 1MO results
of only 1/1000 A. (?V but ?C) At the opposite
extreme, a low voltage in a very low resistance
circuit can produce a very large amount of
current. A 6-v battery connected across a
resistance of 0.01O produces 600A of current. (?V
but ?C)
WK04
2
KEMAHIRAN MANIPULATIF I
EDU3416
The practical fact is that high-voltage circuits
usually do have a small value of current in
electronic equipment. Otherwise, tremendous
amounts of power would be necessary. Example
1. A heater with a resistance of 16O is connected
across the 240v power line. How much is the
current I? I V/R 240/16 15A Example 2. A
small light bulb with resistance of 4800O is
connected across the same 240v power line. How
much is the current I? Although both cases have
the same 240v applied, the current is much less
in Example 2 because of higher resistance.
WK04
3
KEMAHIRAN MANIPULATIF I
EDU3416

• Practice Problems.
• Calculate I for 24v applied across 8O
• Calculate I for 12v applied across 8O
• Calculate I for 24v applied across 12O
• Calculate I for 6v applied across 6O
• INFO
• Transistors and integrated circuits generally
  operate with a dc supply of 3, 5, 6, 9, 12, 20 or
  28v. The current is usually in millionths or
  thousands of one ampere up to about 5A.
• Ohms law states that the current flow through a
  load is proportional to the voltage across the
  load and inversely proportional to the resistance
  of the load when the temperature is constant.

WK04
4
KEMAHIRAN MANIPULATIF I
EDU3416

• The amount of voltage across R must be the same
  as V because the resistance is connected directly
  across the battery.
• Example 3.
• The voltage across the load is 6v which equals to
  the applied voltage.
• V I x R
• 2 x 3 6v
• Practice Problems.
• Calculate V for 0.002A through 1KO
• Calculate V for 0.004A through 1KO
• Calculate V for 0.002A through 2KO

WK04
5
KEMAHIRAN MANIPULATIF I
EDU3416
Continued 4. The I of 6mA flows through a 5KO
R. how much is the IR voltage? 5. How much
current is produced by 60v across 12KO? 6.
Calculate the voltage drop across the load of
470O if the current flow is 25mA. 7. If the
total resistance (R1 R2 R3) of the circuit is
100O, determine the current flowing through
resistor R1 if the voltage across R1 is 10v and
across R2 is 4v if the applied voltage is
25v. 8. A 12v battery is connected in a circuit
having three series-connected resistor having
resistance of 4O, 9O and 11O. Determine the
current flowing through, and the potential
difference across the 9O resistor.
WK04
6
KEMAHIRAN MANIPULATIF I
EDU3416

• Whenever voltage is supplied to a resistive
  circuit, the electric formed will result in the
  movement of electrons. Consequently, the
  collision of these electrons with the atom in the
  resistive material will generate heat. The heat
  generated the power dissipated in resistor is
  symbolized with the alphabet P.
• One watt of power equals the work done in one
  second by one volt of potential difference in
  moving one coulomb of charge. Therefore power in
  watts equals the product of amperes times volts.
• Power in watt, P volts x amperes V x I
• For instance, when a battery produces 2A in a
  circuit, the battery is generating 12w of power.
• Practice Problems.
• A toaster takes 10A from the 240v power line. How
  much power is used?
• How much current flows in the filament of a 300w
  bulb connected to the 240v power line?

WK04
7
KEMAHIRAN MANIPULATIF I
EDU3416
When current flows in a resistance, heat is
produced because friction between the moving free
electrons and the atoms obstructs the path of
electron flow. The heat is evidence that power is
used in producing current. The electric energy
converted to heat is considered to be dissipated
or used up because the calories of heat cannot be
returned to the circuit as electric energy. Power
is also dissipated in resistance of a circuit.
Since power is dissipated is the resistance of
a circuit, it is convenient to express the power
in terms of the resistance R. The V x I formula
can be rearranged as follows
WK04
8
KEMAHIRAN MANIPULATIF I
EDU3416
Practice Problems 1. Current I is 2A in a 5O R.
Calculate P. 2. Calculate the power in a circuit
where the same source of 100v produces 4A in a
25O R. 3. Calculate the power in a circuit
where the source of 100v produces 2A in a 50O R.
4. Resistance R has 10v with 2A. Calculate the
values for P and R. 5. How much current is
needed for a 600w 240v toaster? 6. How much is
the resistance of a 600w 240v toaster? 7. How
much current is needed for a 24O R that
dissipates 600w? 8. The bulb is connected across
the 240v line. Its 300w filament requires current
of 2.5A, calculate the resistance of the filament?
WK04
9
KEMAHIRAN MANIPULATIF I
EDU3416
When the components is a circuit are connected in
successive order with an end of each joined to
an end of the next, they form a series circuit.
The result is only one path for electron flow.
Therefore, the current I is the same in all the
series component.
WK04
10
KEMAHIRAN MANIPULATIF I
EDU3416
The current cannot differ in any way because
there is just one current path for the entire
circuit. The order in which components are
connected in series does not affect the current.
The questions of whether a component is first,
second, or last in a series circuit has no
meaning in terms of current. The reason is that I
is the same amount at the same time in all the
series components. The total resistance RT of a
series string is equal to the sum of the
individual resistances. The string resistances
equals the sum of the individual resistances.
As the entire string is connected across the
voltage source, the current equals the voltage
applied across the entire string divided by the
total series resistance of the string. The
total resistance of a series string equals the
sum of the individual resistances. The formula is
RT R1 R2 R3 .. Rn
WK04
11
KEMAHIRAN MANIPULATIF I
EDU3416
The IR voltage drop across each resistance is
called an IR drop, or voltage drop, because it
reduces the potential difference available for
the remaining resistance in the series circuit.
The whole applied voltage (VT) is equal to the
sum of the individual IR drops. VT V1 V2
V3 Vn Example 1. A voltage source
produces an IR drop of 40v across a 20O R1, 60v
across a 30O R2, and 180v across a 90O R3, all in
series. How much is the applied voltage? VT
V1 V2 V3 40 60 180
280V When an IR voltage drop exists across a
resistance, one end must be either more positive
or more negative than the other end. For either
electron flow or conventional current the actual
polarity of the IR drop is the same.
WK04
12
KEMAHIRAN MANIPULATIF I
EDU3416
The total power used is the sum of the individual
values of power dissipated in each part of the
circuit as a formula, PT P1 P2 P3 ..
Pn Example A voltage source, 60v produces an IR
drop of 20v across a 10O R1, and 40v across a 20O
R2 in series. How much is the total power
dissipated by R1 and R2? P1 V1 x I1 20 x 2
40w P2 V2 x I2 40 x 2 80w PT P1 P2
40 80 120w The total power can also be
calculated as VT x I
WK04
13
KEMAHIRAN MANIPULATIF I
EDU3416
Series-aiding voltages are connected with
polarities that allow current in the same
direction. Voltages are connected series-aiding
when the plus terminal of one is connected to the
negative terminal of the next. Series-opposing
voltages are subtracted. Subtract the smaller
from the larger value, and give the net V the
polarity of the larger voltage. The polarity of
VT is the same as larger V. If two
series-opposing voltages are equal, the net
voltage will be zero. Circuit with voltage
sources in series. Note that V1 and V2 are series
opposing, with to through R1. Their net
effect, then, is 0v. Therefore, VT consists only
V3, equal to 4.5V.
WK04
14
KEMAHIRAN MANIPULATIF I
EDU3416

• An open circuit is a break in the circuit path.
  An open in part results in no current for the
  entire circuit. The resistance of the open is
  very high because an insulator like air takes the
  place of a conducting part of the circuit.
• Practice Problems
• A circuit has 10v applied across a 10O resistance
  R1. How much is the current in the circuit? How
  much resistance R2 must be added in series with
  R1 to reduce the current one-half. Show the
  schematic diagram of the circuit with R1 and R2.
• An R1 of 90kO and an R2 of 10kO are in series
  across a 6v source. Draw the schematic diagram
  and how much is V2?
• Three 20O resistances are in series across a
  voltage source. Show the schematic diagram. If
  the voltage across each resistor is 10v, how much
  is the applied voltage. How much is the current
  in each resistance?

WK04
15
KEMAHIRAN MANIPULATIF I
EDU3416
When two or more components are connected across
one voltage source, they form a parallel circuit.
The applied voltage is the same across
components connected in parallel branches. A
common application of parallel circuits is
typical house wiring to the power line, with many
lights and appliances connected across the 240v
source. For instance, the light bulb is
connected to one outlet and the toaster to
another outlet, but both have the same applied
voltage of 240v.
WK04
16
KEMAHIRAN MANIPULATIF I
EDU3416
In applying Ohms law, it is important to note
that the current equals the voltage applied
across the circuit divided by the resistance
between the two points where that voltage is
applied. Example. How much is the
voltage across R1? How much is I1 through R1? How
much is the voltage across R2? How much is I2
through R2?
WK04
17
KEMAHIRAN MANIPULATIF I
EDU3416

• Example.
• An R1 of 20O, an R2 of 40O and R3 of 60O are
  connected in parallel across the 240v power line.
  How much is the total line current IT?
• Current I1 for the R1 branch is 12A and I2 is 6A
  and I3 is 4A. The total current in the main line
  is 11A.
• Practice Problems.
• Two branches R1 and R2 across 240v power line
  draw a total line current IT of 15A. The R1
  branch takes 10A. How much is the current I2 in
  the R2 branch?
• Three parallel branch current are 0.1A, 500mA and
  800µA. Calculate IT.
• Parallel branch currents are 1A for I1, 2A for I2
  and for I3. Calculate IT.
• Assume IT 6A for three branch currents, I1 is
  1A and I2 is 2A. Calculate I3.

WK04
18
KEMAHIRAN MANIPULATIF I
EDU3416
The total resistance across the main line in a
parallel circuit can be found by Ohms law.
The parallel resistance of R1 and R2
indicated by the combined resistance RT, is the
opposition to the total current in the main line.
The total load connected to the source voltage
is the same as though one equivalent resistance
of 20O were connected across the main line.
Example. Two branches, each with a 5A current,
are combined across a 90v source. How much is the
equivalent total resistance RT?
WK04
19
KEMAHIRAN MANIPULATIF I
EDU3416
Reciprocal resistance formula can be derive from
the fact that IT is the sum all of the branch
currents or IT I1 I2 I3 . In
When there are two parallel resistance and
they are not equal, it is usually quicker to
calculate the combined resistance by the
following method.
WK04
20
KEMAHIRAN MANIPULATIF I
EDU3416
Special rules can help in reducing parallel
branches to a simpler equivalent circuit.
R1 and R4 are equal and in parallel, 60O.
Therefore, they are equivalent to the 30O R14.
Similarly, R2 and R3 are equivalent to the 10O of
R23.
WK04
21
KEMAHIRAN MANIPULATIF I
EDU3416
Practices Problems. 1. Find RT for three 4.7MO
resistances in parallel. 2. Find RT for 3MO with
2MO. 3. Find RT for two 20O resistances in
parallel with 10O
WK04
22
KEMAHIRAN MANIPULATIF I
EDU3416
The total power equals the sum of the individual
values of power in each branch since the power
dissipated in the branch resistances must come
from the voltage source. Example. The
branch current I1 then is VA/R1 which equals 1A.
Similarly, I1 is 10/5 or 2A. The total IT is 1
2 3A. The power dissipated in each branch R
is VA x I. In the R1 branch, P1 is VA x I1 or
10w. For the R2 branch, P2 is VA x I2 20w. PT
P1 P2 10 20 30w.
WK03
23
KEMAHIRAN MANIPULATIF I
EDU3416

• Practice Problems.
• Two parallel branches each have 2A at 240V.
  Calculate PT.
• Three parallel branches of 10, 20 and 30O have
  60V applied. Calculate PT.

WK03
24
KEMAHIRAN MANIPULATIF I
EDU3416
In many circuits, some components are connected
in series to have the same current, while others
are in parallel for the same voltage. Such a
circuit is called series-parallel circuits used
to provide different amounts of current and
voltage with one source of applied voltage. In
order to calculate current and voltage in
series-parallel resistor circuit, it is very
important to understand the concept of the series
and parallel resistor circuit. Example
WK03
25
KEMAHIRAN MANIPULATIF I
EDU3416
Solution
WK03
26
KEMAHIRAN MANIPULATIF I
EDU3416
Practice Problems
WK03