 CHAPTER 4 OHM'S LAW CURRENT-RESISTANCE-VOLTAGE You learned in chapter 1 that the flow of CURRENT is influenced by both VOLTAGE and RESISTANCE. INCREASE the VOLTAGE, and you also INCREASE the CURRENT. But if you INCREASE the RESISTANCE, the CURRENT DECREASES. In some respects, the flow of current in a circuit is like the movement of water in a pipe. The rate of water flow is influenced by the size or the pipe and by the amount of pressure applied to the water. Naturally a large pipe with few obstacles will allow more water to move through it than a smaller one. And if a high pressure is applied a greater flow will take place than if the pressure is lower. Don't become confused by thinking that voltage and water pressure are exactly the same thing. While water pressure has an effect on the movement of water similar to the effect of voltage on the movement of electrons, they are not the same. Water pressure is a "push" from 33 behind, while electrons are "pulled" toward the higher voltage. The results of the two are much the same, but you should keep the difference in mind. The relationship of the CURRENT to both VOLTAGE and RESISTANCE is given in Ohm's Law. It reads, "The current flowing in a circuit VARIES DIRECTLY as the VOLTAGE and INVERSELY as the RESISTANCE." Since Ohm's Law states the relationship as a PROPORTION, you may write the law in a simple formula- Current (I) = Voltage (E) / Resistance (R) or just I = E / R FINDING THE CURRENT BY USING OHM'S LAW Ohm's Law provides a simple method for finding the CURRENT flowing in a circuit if you know the VOLTAGE and RESISTANCE. Figure 28.-Simple circuit. In figure 28, a lamp is connected to a battery with an emf of 1.5 volts. The resistance of the lamp is 15 ohms. To find the current flowing, substitute the values of E and R in the formula of Ohm's Law and solve- 34 I = E / R = (1.5 / 15) = 0.10 amperes Thus whenever you know the applied VOLTAGE and the RESISTANCE of a circuit, you can find the current by using Ohm's Law. FIND THE RESISTANCE BY USING OHM'S LAW By performing an easy mathematical maneuver, Ohm's Law can be changed- from I = E / R to R = E / I In this new form you use Ohm's Law to find the RESISTANCE of a circuit if you know the CURRENT and the APPLIED VOLTAGE. The ammeter, in figure 29, indicates a current of 0.02 amperes flowing through the circuit. The applied E is Figure 29.-Simple circuit. three volts. To find the resistance of the circuit, substitute in the values of E and I in the formula- R = E / I = 3/0.02 = 150 ohms 35 FINDING THE VOLTAGE BY USING OHM'S LAW When a current flows through a resistance, a DROP in voltage occurs. The LARGER the current and the HIGHER the resistance, the GREATER will be the DROP in voltage. To find this voltage drop, Ohm's Law is again changed from I = E / R to E = IR The voltage in the new formula is equal to the PRODUCT of the current and voltage. Hence if you know the resistance, and the current flowing through it, you can find the VOLTAGE DROP ACROSS THE RESISTANCE. Voltage drops across resistances are usually called "IR" drops. Figure 30.-Simple circuit. The battery in figure 30 is forcing a current of 0.02 amperes to flow through a lamp whose resistance is 100 ohms. The IR drop across the lamp will be- E = IR = 0.02 X 100 = 2 volts IR DROPS ABOUT A CIRCUIT It is not unusual for a circuit to contain several resistances in series. Since the same current flows through all the resistances, IR drop will be present across each resistor. 36 Figure 31.-IR drops about a series circuit.
Figure 31 is a basic filament circuit used in some small AC-DC receivers. The filaments of the 5 tubes T1 ..... T5 are connected in series with an outside resistor R. The resistance of each filament is given below the tubes. An ammeter in the circuit indicates a current of 0.15 amperes to be flowing. The IR drop across each tube and resistor will be equal to-

resistance X 0.15

So for the series circuit the IR drops will be-

 R - 80 X .15 = 12.00 volts T1 - 94 X .15 = 14.00 volts T2 - 80 X .15 = 12.00 volts T3 - 80 X .15 = 12.00 volts T4 - 233 X .15 = 34.95 volts T5 - 233 X .15 = 34.95 volts Total 119.90 volts

All the IR drops added together equal 119.90 volts. Now look at the applied voltage-120 volts. It suggests that the sum of the IR drops about a closed circuit are EQUAL to the APPLIED E.

What happened to the other 0.10 of a volt (120 - 119.90)? It's distributed over the connecting wires and the meter.

Hence, in any circuit, "The sum of the individual IR drops is EQUAL TO the APPLIED VOLTAGE." That statement is known as Kirchhoff's First Law.

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Kirchhoff also discovered a law about current. It states, "As much current flows AWAY from a point as INTO it." Figure 32.-Current flowing through a divided circuit. The battery in figure 32 causes a current of 2 amperes to flow through the circuit. At point A, the circuit divides and 1 ampere flows through each leg. At point B, the circuit branches again with 1/2 ampere flowing through each branch. At point C, the three branches come together, and 2 amperes flow away. In the case of each of the points, A, B, and C, AS MUCH CURRENT FLOWS AWAY AS INTO THE POINT. While this law may sound simple, it is one of the most useful in the study of electricity. POWER POWER is the RATE OF DOING WORK. Suppose you are assigned to help load ammunition. If you have a liberty coming up when the loading is done, you will work hard to get the job done in a hurry. But if you are not going any place, chances are you will do the job in a more leisurely manner. In the first case, the RATE of doing work is high. You will use up a lot of energy in a short period of time. In other words, a lot of POWER will be exerted in doing the 38 job. But when you go about your job leisurely, energy is being expended at a slow rate and the power consumption is low. In the same way, when a lot of electrons are forced through a resistor in a short period of time, a lot of energy is used and the power consumption is great. If fewer electrons are forced through the same resistor, less power is dissipated. Actually, the amount of power consumed in forcing electrons through a resistor is proportional to the SQUARE of the CURRENT times the RESISTANCE, or- Power (in watts) = I2R Thus, if 0.02 amperes is flowing through 1,000 ohms of resistance, the power being dissipated will be Watts = .022 X 1,000 Watts = 0.4 But if the current is increased to .04 amperes, the power goes up to- Watts = .042 X 1,000 Watts = 1.6 There are two other formulas that you can use to find the power of a circuit. One is- Watts = E X I and the other is- Watts = E2 / R You may use any of these three formulas to find the power. They all express rate of doing work. 39