# Kirchhoff s Laws and Resistance in Visual Studio .NET Encoder Code 128 Code Set A in Visual Studio .NET Kirchhoff s Laws and Resistance

Kirchhoff s Laws and Resistance
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Fig 2-7 A schematic representation of a resistor
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With the equation in this form, we see that for a given applied voltage, if the resistance of the material is larger, the resulting current will be smaller The inverse of resistance is the conductance, G G= 1 (siemens) R (28)
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As indicated, the SI unit of conductance is the siemens, a name that comes from a mysterious German scientist who studied electric properties of materials some time ago While we will stick to SI units in this book, be aware that conductivity is also measured in mhos, which are denoted by an upside-down omega symbol We indicate a resistor in a circuit by drawing a jagged line, as shown in Fig 2-7 If a device resists the ow of current, the energy has to go somewhere This is usually re ected in the emission of heat or light from the device Resistance is found in many practical electric components and appliances Perhaps the most familiar example of a resistor is the lament in a light bulb, where the resistance gives rise to light Another example is a toaster, where resistive elements give off heat and some light that is useful to toast bread In other cases, resistance might not be as useful; for example, an electric chord might have a bit of resistance that results in heat
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Power in a Resistor
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The power absorbed or delivered by a resistor can be calculated from the expression P = VI together with Ohm s law (26) V = RI If we know the resistance and the voltage, then P = VI = V2 = GV 2 R (29)
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On the other hand, if we know the current through the resistor then we can write P = VI = RI 2 (210)
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Remember, a resistor is an element that gives off energy, usually in the form of heat and sometimes in the form of light Hence, a resistor always absorbs power
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When doing circuit analysis with a network that contains resistors, we apply KCL and KVL using Ohm s law to relate the voltage to the current as necessary The best way to proceed is to look at some examples EXAMPLE 2-5 Find the three unknown currents shown in Fig 2-8 SOLUTION We will denote the voltage across each resistor R by VR First we apply KVL to each of the two panes or loops in the circuit Going in a clockwise direction, the loop on the left-hand side of Fig 2-8 gives 7 + V5 = 0 where V5 is the voltage across the 5 resistor in the center We conclude from KVL that V5 = 7 V Using Ohm s law (26) we can nd the current through the
I3 I2
+ +
10
20 V
Fig 2-8 Circuit analyzed in Example 2-5 Note the voltage polarities that have been speci ed for each resistor
Kirchhoff s Laws and Resistance
resistor, which is just I2 I2 = V5 7 = = 14 A 5 5 (211)
Next we apply KVL to the right-hand loop in Fig 2-8 Again, we take the loop in a clockwise direction This gives V5 + V10 + 20 + V3 = 0 (212)
Although we know V5 = 7 V this equation leaves us with two unknowns , With one equation and two unknowns we need more information to solve the problem Some extra information comes in the form of Ohm s law The same current I3 ows through the 10 and 3 resistors Hence V10 = 10I3 and V3 = 3I3 and we can write (212) as 7 + 10I3 + 20 + 3I3 = 0 I3 = 1 A Knowing two of the currents, we can solve for the other current by considering KCL at the top-center node We take + for currents entering the node and for currents leaving the node This gives I1 I2 I3 = 0 Therefore we have I1 = I2 + I3 = 14 1 = 04 A Now we apply Ohm s law again to get the voltages across each of the resistors V10 = 10I3 = 10 ( 1) = 10 V V3 = 3I3 = 3 ( 1) = 3 V Notice the minus signs These tell us that the actual voltages have the polarity opposite to that indicated in Fig 2-8 EXAMPLE 2-6 Let s consider a simple abstract model of a toaster Our model will consist of the wall outlet, an electric chord, a switch, and a heating element The wall outlet