IMG_20191022_095827.jpg - 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

IMG_20191022_095827.jpg - OHM'S LAW in the hose the...

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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 Intentiontar...
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  • Fall '19
  • SK Abid

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