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**Unformatted text preview: **OHM'S LAW
in the hose, the pressure valve as the applied voltage, and the size of the
hose as the factor that determines the resistance. If the pressure valve;
Since the battery in Fig. 4:2 is connected directly across the resistor,
closed, the water simply sits in the hose without a general direction, mud
the voltage VR across the resistor must be equal to that of the supply. Ap-
like the oscillating electrons in a conductor without an applied voltage
plying Ohm's law, we obtain
When we open the pressure valve, water will flow through the hose much
VR _ E
like the electrons in a copper wire when the voltage is applied. In other
R R
words, the absence of the "pressure" in one case and the voltage in the other
simply results in a system without direction or reaction. The rate at which
Note in Fig. 4.2 that the voltage source "pressures" current (conven-
the water will flow in the hose is a function of the size of the hose. A hose
tional current) in a direction that leaves the positive terminal of the
with a very small diameter will limit the rate at which water can flow
supply and returns to the negative terminal of the battery. This will
through the hose, just as a copper wire with a small diameter will have a
always be the case for single-source networks. (The effect of more than
high resistance and will limit the current.
one source in the same network is investigated in a later chapter.) Note
In summary, therefore, the absence of an applied "pressure" such as
also that the current enters the positive terminal and leaves the negative
voltage in an electric circuit will result in no reaction in the system and
terminal for the load resistor R.
omra
For any resistor, in any network, the direction of current through a
(b)
FIG. 4.1
no current in the electric circuit. Current is a reaction to the applied volt.
George Simon Ohm.
age and not the factor that gets the system in motion. To continue with
resistor will define the polarity of the voltage drop across the resistor
Courtesy of the Smithsonian
the analogy, the greater the pressure at the spigot, the greater is the rate
as shown in Fig. 4.3 for two directions of current. Polarities as estab-
FIG. 4.3
Institution. Photo No. 51,145.
of water flow through the hose, just as applying a higher voltage to the
lished by current direction become increasingly important in the analy-
Defining polarities.
German (Erlangen, Cologne)
same circuit results in a higher current.
(1789-1854)
Substituting the terms introduced above into Eq. (4.1) results in
ses to follow.
Physicist and Mathematician
Professor of Physics, University of Cologne
Current - potential difference
EXAMPLE 4.1 Determine the current resulting from the application of
In 1827, developed one of the most important laws
resistance
a 9 V battery across a network with a resistance of 2.2 0.
of electric circuits: Ohm's law. When the law was
first Introduced, the supporting documentation was
Solution : Eq. (4.2):
considered lacking and foolish, causing him to lose
his teaching position
and
search for a living doing
(amperes, A)
(4.2)
VR _ E 9V
odd jobsand
RR 220 = 4.09 A
das a major c
Eq. (4.2) is known as Ohm's law in honor of Georg Simon Ohm (Fig. 4.1).
The law states that for a fixed resistance, the greater the voltage (or pres-
EXAMPLE 4.2 Calculate the resistance of a 60 W bulb if a current of
sure) across a resistor, the greater is the current; and the greater the re-
500 mA results from an applied voltage of 120 V.
sistance for the same voltage, the lower is the current. In other words, the
current is proportional to the applied voltage and inversely proportional
Solution : Eq. (4.4):
to the resistance.
R =
120 V
By simple mathematical manipulations, the voltage and resistance
= 7 500 X 10-3 = 240 12
can be found in terms of the other two quantities:
16 V
EXAMPLE 4.3 Calculate the current through the 2 kf2 resistor in
2 kn
E = IR
( volts, V)
(4.3)
Fig. 4.4 if the voltage drop across it is 16 V.
FIG. 4.4
Solution :
Example 4. 3.
and
(ohms, )
(4.4)
_V_ 16V
1= R 2X 1030
= 8 MA
All the quantities of Eq. (4.2) appear in the simple electrical circuit in
Fig. 4.2. A resistor has been connected directly across a battery to estab
EXAMPLE 4.4 Calculate the voltage that must be applied across the
lish a current through the resistor and supply. Note that
soldering iron in Fig. 4.5 to establish a current of 1.5 A through the iron
the symbol E is applied to all sources of voltage
if its internal resistance is 80 0.
1 = 1.5 A
and
Solution:
E = VR = IR = (1.5 A)(80 0) = 120 V
R
the symbol V is applied to all voltage drops across components of the
network.
FIG. 4.2
Basic circuit.
Both are measured in volts and can be applied interchangeably in
In a number of the examples in this chapter, such as Example 4.4,
FIG. 4.5
Egs. (4.2) through (4.4).
the voltage applied is actually that obtained from an ac outlet in the
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- Fall '19
- SK Abid