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Metric Prefixes 
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METRIC PREFIXES Numerical values in electronics circuits range from very small to very large. A resistor may be in the millions of ohms and a current might be in the millionths of an ampere. For that reason several methods are available to help express the very small or very large number. Scientific notation is a form that expresses numbers as some value 1or higher but less than 10, times some power of 10. (Examples: 91,200 = 9.12 x 10^{4 }and 0.000045 = 4.5 x 10^{5}) A scientific calculator will express numbers this way when a very large or very small result is produced from a calculation. Another more popular form used in technical fields is referred to as engineering notation. The engineering form expresses numbers as some value 1 or higher but less than 1000, times a power of 10 that is a multiple or submultiple of 3. (Examples: 91,200 = 91.2 x 10^{3 } and 0.000045 = 45 x 10^{6}) The metric or preferred decimal multiplier format follows the engineering concept but uses a metric term to express the power of 3. (Examples: 91.2 x 10^{3} = 91.2k and 45 x 10^{6 }= 45u) Metric terms make it easier to write and speak powers of 10 in engineering form. Instead of having to list a resistor value as 47,000 W, we would write it as 47 kW, or a current of 0.0025 amps as 2.5 mA. Because so many component values are either fractional parts or multiples of their base unit definitions, the metric prefix is the norm in specifying electronic parts. The same is true in expressing circuit parameters and measurements. The more common metric terms are listed in the table below. 
Prefix 
Symbol 
Power of Ten 
Standard Expression 
pico  p  10^{12}  0.000000000001 
nano  n  10^{9}  0.000000001 
micro  u  10^{6}  0.000001 
milli  m  10^{3}  0.001 
units  not applicable  10^{0}  1 
kilo  k  10^{3}  1,000 
mega  M  10^{6}  1,000,000 
giga  G  10^{9}  1,000,000,000 
tera  T  10^{12}  1,000,000,000,000 
Often it is required to express or interpret a value in a different term than what is already given. A meter reading range may give a measurement using one prefix but the lab manual asks that the result be recorded in another or the calculator gives the result in a power of three different than the choices on a test. From the chart above, realize that as we go down the list the terms are increasing by a thousand. Pico is a very small unit prefix, and the next one, nano, is a thousand times bigger. That means that one nano is a thousand picos or that one pico is one thousandth (1/1000) of a nano. So to express a value in the next larger unit divide by 1,000 which is the same as moving the decimal point three place to the left. To express a value in the next lower unit multiply by a 1,000 which is the same as moving the decimal three places to the right. As an example, a current of 3.5 mA could be expressed as 3,500 uA (moving the decimal 3 places to the right) as the next smaller unit or as 0.0035 A (moving the decimal 3 places to the left) as the next larger prefix. Just remember that for each unit change move the decimal three places and if going to a larger unit move it left and if going to a smaller unit move it right. 
I = P/V = 3 W/12 V = 0.25 A or 250 mA R = V^{2}/P = (12^{ }V)^{2 }/3 W= 144/3 = 48 W R = V/I = 12 V/0.025 A = 48 W I = V/R = 12V/48 W = 0.25 A or 250 mA 
The simple circuit is the simplest of all circuits to solve. In the circuit shown, the source, 12 V, is given and the load is a lamp rated to use power at the rate of 3 watts when connected to a 12 V source. Generally, that is the way lamps are rated. Since V and P are given, the circuit can be solved several different ways. Both I and R can be solved using the appropriate formula. See the solution at right. Another possibility would have been to find I first and then find R = V/I. Still another would have been to find R first, and then find I = V/R.

Note the direction of current in the series circuit. Using the electron flow model, Current (I) flows from the negative terminal of the source through each resistor back into the positive terminal of the source. This I flow creates voltage drops across each resistor. According to polarity assignment rules, where I enters a component is negative and where it leaves is positive. Note in the circuit that the top of each resistor is positive with respect to the bottom. This would be very important to know when measuring the voltage drops. The positive (+) lead of the voltmeter should be placed on the positive points and the negative lead () on the negative points. 
Given the circuit shown, note the only given values are source voltage and the load resistances. To solve this circuit first find R_{T}. R_{T} = R_{1} + R_{2} + R_{3} = 500 W + 1500 W + 1000 W = 3000 W or 3 kW Find the circuit current. I_{T }= V/R_{T} = 12V/3000 = 0.004A or 4mA Find the voltage drop across each resistor. (I is the same in all parts) V_{1} = IR_{1} = 0.004 A x 500 W = 2 V V_{2 } = IR_{2} = 0.004 A x 1500 W = 6 V V_{3} = IR_{3} = 0.004 A x 1000 W = 4 V As a check, the sum of the voltage drops should equal the source voltage. V_{1}+V_{2}+V_{3 }= 2 V + 6 V + 4 V = 12 V which is the same as Vs. Last, solve for power in each load and total power for the circuit. Since we know I, V, and R, the load powers can be found by any of the power equations. They all should give the value, the simplest one is P = IV. P_{1} = IV_{1} = 0.004 A x 2 V = 0.008 W or 8 mW P_{2} = IV_{2} = 0.004 A x 6 V = 0.024 W or 24 mW P_{3} = IV_{3} = 0.004 A x 4 V = 0.016 W or 16 mW P_{T} = IV_{S} = 0.004 A x 12 V = 0.048 W or 48 mW As a check, the sum of the load powers should equal the total power. P_{T} = P_{1} + P_{2} + P_{3} = 0.008 W + 0.024 W + 0.016 W = 0.048 W or 48 mW, which does agree. This circuit has been totally solved. We know V, I, R, P and polarity assignment for each load and V_{s}, I, R_{T}, P_{T} and current direction for the circuit from the source perspective. Note that the largest voltage drop and largest power is in the largest resistor, R_{2}. 
Sources are often wired in series to get a higher voltage than a single source acting alone. Sources connected this way are said to be seriesaiding. In the circuit shown, the lamp voltage rating is 12 volts. If the voltage is much higher than this then the lamp would burn out. If the voltage was lower, then the lamp would dim or maybe not shine at all. Note in this circuit that two 6 volt sources are connected endtoend or seriesaiding to obtain the 12 volts needed by the lamp. Just remember that the current in series is the same through all parts, so the sources need to have similar current ratings. 
I = P/V = 10 W/12 V = 0.833 A or 833 mA 
Seriesaiding sources are connected endtoend, positivetonegative, and the total voltage is the sum of all the individual values. This is commonly done with batteries to increase the total power without decreasing the operating life of the circuit. A batteries life is rated as to how much current it can supply for a period of time. If we want to keep current flow low but still increase power we can increase the voltage (Remember that P = I V). Flashlights are good examples. They all use 1.5 V cells, but the ones with more cells are brighter. In the circuit shown, the current will be 0.5 amps. If we wanted to make a two cell flashlight that would operate for the same length of time, the current would have to remain at 0.5 amps and the lamp would have to be designed to operate with 3 volts. The power from that lamp would be P = IV = 0.5 A x 3 V = 1.5 W, which is only half as intense as the four cell flashlight. 
I_{L1 }= P/V = 3 W/6V = 0.5 A or 500 mA What would happen in this circuit if one of the cells was near full discharge? Remember that the current in series has to be the same through all parts. If one cell has diminished capacity, it limits the current for the entire circuit. The lamp would be very dim or not shine at all. 
Sources can be connected as seriesopposing which means that some cancel out part of the others. In other words they are trying to send current in opposite directions. In a battery symbol, the end with the shorter line is negative and by the electron flow model, current flows from the negative terminal of the source to the positive terminal. To find the net voltage of all the sources, determine which way each are trying to send current around a circuit, clockwise or counterclockwise. Add all those for each direction and then subtract the two. This will be the net source voltage. The current direction for the circuit will be determined by the larger of the two sums. Seriesopposing sources is rarely done on purpose but it does often happen by accident, as when someone installs one of several series batteries backward. In the circuit shown note that for V1, V2, and V4 the current direction would be clockwise around the circuit, but for V3 it would be counterclockwise. Here V1, V2, and V4 are aiding each other but are being opposed by V3. V1+V2+V4 = 1.5V + 1.5V + 1.5V = 4.5 V. The net voltage is 4.5 V  V3 = 4.5V  1.5V = 3 V. The current would flow clockwise around the circuit. Note that this the same flashlight circuit from above but with one cell installed backward. The resistance of the lamp can be found from the rated lamp values as R = V^{2}/P = 6^{2}/3 = 36/3 = 12 W. With only 3 volts, the current would be cut in half (I = V/R = 3/12 = 0.25 A) and since both V and I are now onehalf, the actual power would only be onefourth the rated value (P = IV = 0.25 x 3 = 0.75 W). The light would not shine.  A version of Kirchhoff's Voltage Law can also be used to find the net voltage. This law allows us to find the difference in potential between two points and the polarity at the initial point with respect to the end point. As we move from the starting point through the components toward our end point we use the sign as we enter a component and the voltage across that component. We then add these algebraically to get the voltage difference. If in the circuit above we start to the left of V1 and move to the right of V4, we get +V1 + V2  V3 + V4. This would give us +3 volts. This means that the net voltage from all four cells is 3 volts with the left side positive and the right side negative. 
Kirchhoff's Laws are an extension of Ohm's Law. The two laws, one for voltage and one for current, are very helpful in solving and analyzing circuits, especially the more complex seriesparallel circuits. Kirchhoff's Voltage Law can be stated as "around any closed loop, the net source voltage equals the sum of the voltage drops" or "around any closed loop, the algebraic sum of all voltages is zero". These would be expressed as V_{T} = V_{1} + V_{2} + ^{. . . .} + V_{N }or V_{T }  V_{1 } V_{2}  ^{. . .}  V_{N} = 0. They are the same mathematically. Kirchhoff's Current Law can be stated as "the sum of the currents entering a point must equal the sum of the currents leaving that point". One expression is I_{T} = I_{1} + I_{2} + ^{. . . }+ I_{N} 
Note that in a parallel circuit, each load has connections directly to the source. That is why each has the same voltage Also each load offers a separate path for current to flow. That is why the loads could be operated independently of each other if each had its own switch. Turning each load on or off would not prevent the other from still having the ability to operate because it would still have a connection to the source. The same would be true if one component were to open due to a defect. The others could still operate normally. This is why in your home if one light bulb burns out the others stay on. In the circuit above, the one switch is in the part of the circuit where the source current flows so it could be used to turn on or off the entire circuit.

PARALLEL CIRCUIT SOLUTION Given the circuit shown, note the only given values are source voltage and the load resistances. The simplest way to solve this circuit is to find the branch currents first. Note that V is the same across all parts in a parallel circuit. I_{1} = V/R_{1} = 12 V/500 W = 0.024 A or 24 mA I_{2} = V/R_{2} = 12 V/1500 W = 0.008 A or 8 mA I_{3} = V/R_{3 }= 12 V/1000 W = 0.012 A or 12 mA Find total current as the sum of the branch currents. I_{T} = I_{1} + I_{2}+ I_{3} = 0.024 A + 0.008 A + 0.012 A = 0.044 A or 44 mA Find the total resistance. R_{T} = V/I_{T} = 12 V/0.044 = 273 W Note: R_{T} can also be found by the reciprocal formula for any number of parallel resistors and should give the same result. R_{T} = 1/(1/R1 + 1/R2 + 1/R3) = 1(1/500 + 1/1500 + 1/1000) = 1/(0.002 + 0.00067 + 0.001) = 1/0.00367 = 273 W Solve for power in each load and total power for the circuit. Since we know I, V, and R, the load powers can be found by any of the power equations. They all should give the value, the simplest one is P = IV. P_{1} = I_{1}V = 0.024 A x 12 V = 0.288 W or 288 mW P_{2} = I_{2}V = 0.008 A x 12 V = 0.096 W or 96 mW P_{3} = I_{3}V = 0.012 A x 12 V = 0.144 W or 144 mW P_{T} = I_{T}V = 0.044 A x 12 V = 0.528 W or 528 mW As a check, the sum of the load powers should equal the total power. P_{T} = P_{1} + P_{2} + P_{3} = 0.288 W + 0.096 W + 0.144 W = 0.528 W or 528 mW, which does agree.

The two reasons for connecting sources in parallel is to deliver more current to a load than what a single source acting alone could deliver, or to operate a given load for a longer time interval before having to recharge or replace the batteries. Any power source will have a rated maximum current that it can deliver and still maintain the specified voltage. If the current draw exceeds that limit then the source may blow a fuse, go into a current limit mode and shut down, or even possibly become damaged. For this reason two or more sources could be wired in parallel and the total load current would be split between the sources. In the circuit shown, the lamp will try to draw 5 amps from a 12 volt source. Assume that the maximum rated current for each source is 3 amps. If the circuit was operated with either source acting alone, it would be overloaded. If both are connected the total current would be split between them and each would supply 2.5 amps which is well below their maximum I rating. Note: it is extremely important that these sources have the same voltage rating. The other reason for connecting sources in parallel is to extend the operating life when using batteries as sources. Battery capacity is generally given as an amount of current that can be delivered for a period of time in hours. The greater the current the sooner it becomes discharged. This rating is expressed in amperehours, which is the product of amps and hours. If the battery capacity in amperehours and the load current is known then the operating life of the circuit can be found. As an example, if a battery was rated at 50 amphours, it could deliver 1 A for 50 hours (1 A x 50 hrs = 50 Ahr). If the load current increased to 5 amps then the time would decrease to 10 hours (5 A x 10 hrs = 50 Ahr). In the circuit shown, assume that the batteries are rated at 20 Ahr each and that the lamp is part of an emergency lighting system. With only one battery source the light could only be expected to shine for 20 Ahr/5 A or 4 hours. With both batteries, each is now only supplying 2.5 amps, so the operating time would be double since 20 Ahr/2.5 A = 8 hours. 
I_{L1}= P/V = 60W/12V = 5A S1 closed & S2 open, I_{V1} = 5 A & I_{V2} = 0 A S1 open & S2 closed, I_{V1} = 0 A & I_{V2} = 5 A S1 & S2 closed, I_{V1} = 2.5 A & I_{V2} = 2.5 A 

SeriesPARALLEL CIRCUIT # 1 SOLUTION In electronic circuits more complex arrangements are the norm. Circuits were all of the components are neither in series or parallel are called seriesparallel circuits. This more complex circuit will have some of the properties of both series and parallel circuits. In other words, there will be parts with different voltages and parts with different currents. The rules for solving these circuits are the same as for the previous circuits, just apply the series rules for the series components and parallel rules for parallel components. These circuits can be very complex with many different voltage divisions and current divisions. The one shown to the right and the one in the next frame are as simple as these circuits can be. Often to solve these circuits, requires redrawing the circuit to better see how parts are related, and then simplifying the circuit a few steps at a time. In the original circuit, top right, it can be seen that current flows from the negative end of the source (the bottom) counterclockwise around the circuit. It would flow up through R_{3} and then divide into I_{1} through R_{1} and I_{2} through R_{2, }and then recombine into I_{T} at the top before returning to the source. This makes R_{1} and R_{2} in parallel with each other. The total current (I_{T}) and I_{3} would be the same and according to Kirchhoff's Current Law, I_{1} + I_{2} = I_{T. }So this circuit has three different current values. Note since R_{1} and R_{2} are in parallel they would have the same voltage, but different currents. To solve first find the equivalent resistance of R_{1} and R_{2} in parallel. Req = 1/(1/1200 + 1/2400) = 800 W. Now we can redraw the circuit and see that this R1R2 equivalent is in series with R_{3}. R_{T} then is the sum of these two values. R_{T} = 800 + 1200 = 2000 W. I_{T }= V/R_{T} = 12/2000 = 0.006 A or 6 mA. The simplified circuit shows that there will be two voltage drops, one across the parallel combination and one across R_{3}. These can be found as V_{eq} = IR_{eq} = 0.006 x 800 = 4.8 V and V_{3} = IR_{3} = 0.006 x 1200 = 7.2 V. Kirchhoff's Voltage Law says that the two drops should add to equal V_{S} and 4.8 V + 7.2 V = 12 V. So far so good. Now that we know that V_{1} and V_{2} = 4.8 V, I_{1} and I_{2} can be found. I_{1} = V_{1}/R_{1} = 4.8/1200 = 0.004 A or 4 mA, and I_{2 } = V_{2}/R_{2} = 4.8/2400 = 0.002 or 2 mA. Kirchhoff's Current Law says that these two currents should add to equal the current that they split from and 2 mA + 4 mA = 6 mA. We can plug all the values back in to the original circuit to see the solution. Now that we know V, I, and R for each part we can also solve for part powers and total power. 

SeriesPARALLEL CIRCUIT # 2 SOLUTION
This circuit is solved similar to the last. Source current (I_{T}) would flow from the negative end of the source and then split into two different currents at the lower junction point. One of these branch currents would flow up through R_{2} and then R_{1}. The other would flow up through R3. These two would recombine at the top junction point becoming I_{T} returning to the source. Note that R_{3} has connections directly to the source. So it is a branch totally independent of the other. V_{3} would be the same as the source voltage or 12V. If V_{3} = 12V then I_{3} = V_{3}/R_{3} = 12/1200 = 0.01 A or 10 mA. Since the same current flows through R_{1} and R_{2}, that makes these two in series. R_{1} and R_{2} can be added to get the total resistance in that branch (we will call it R_{1&2}). R_{1} + R_{2} = 1200 + 2400 = 3600 W. Now that we know the branch resistance we can find the branch current, I_{1&2} = V/R_{1&2} = 12/3600 = 0.00333 A or 3.33 mA. Now that we know the current flowing in the R_{1&2} branch we can go back to the original circuit and solve for V_{1} and V_{2}. Since they both have the same current, solve for V_{1} = I_{1&2} x R_{1} = 0.00333 x 1200 = 4 V and V_{2} = I_{1&2} x R_{2} = 0.00333 x 2400 = 8V. Now that we know the two branch currents we can solve for the source parameters two different ways. First way Find I_{T} by adding the two branch currents and then find R_{T} as V/I_{T}. I_{T} = I_{1&2} + I_{3} = 0.00333 A + 0.01 A = 0.01333 A or 13.33 mA R_{T} = 12V/0.01333A = 900W Second way  Find R_{T} by solving for the two parallel resistors in the simplified circuit and then find I_{T} = V/R_{T}. R_{T} = 1/(1/R_{1&2} + 1/R_{3}) = 1/(1/3600 + 1/1200) = 900W Then I_{T} = V/R_{T} = 12/900 = 0.01333 A or 13.33 mA. Note that both methods give the same result. We now know every voltage and current for each load as well as the source. Power for each part and for the source can now be found using any of the power formulas. 
This concludes the introduction to electronics study material. This material illustrates the type of material that a beginning student in electronics at NWSCC would cover in ILT101  Survey of Electronics and ILT138  DC Fundamentals. DC Fundamentals has a corequisite lab class, ILT139  DC Fundamentals Lab, in which students get to prove circuit theory through handson laboratory exercises. For more information about this program of study see Electronics at NWSCC or contact the college. Thanks for staying to the end. 
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