Kirchhoff s Laws and Resistance in Visual Studio .NET
Kirchhoff s Laws and Resistance Code 128 Code Set A Creation In VS .NET Using Barcode encoder for .NET Control to generate, create Code 128 Code Set B image in .NET applications. Code128 Recognizer In .NET Framework Using Barcode recognizer for .NET Control to read, scan read, scan image in VS .NET applications. Fig 27 A schematic representation of a resistor
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Create Data Matrix 2d Barcode In None Using Barcode creation for Software Control to generate, create Data Matrix 2d barcode image in Software applications. Decode Code 3/9 In Java Using Barcode reader for Java Control to read, scan read, scan image in Java applications. 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 25 Find the three unknown currents shown in Fig 28 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 lefthand side of Fig 28 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 28 Circuit analyzed in Example 25 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 righthand loop in Fig 28 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 topcenter 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 28 EXAMPLE 26 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

