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Fig 2-5 Circuit solved in Example 2-3
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Circuit Analysis Demysti ed
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Fig 2-6 The three-loop circuit of Example 2-4
Solving for the unknown voltage Vx = 5 20 + 12 + 18 = 5 V EXAMPLE 2-4 Consider the circuit shown in Fig 2-6 Find the unknown voltages SOLUTION To be consistent and therefore reduce our chances for error, we again consider clockwise loops, taking each voltage drop we encounter to be positive and each voltage rise we encounter to be negative There are three loops in the circuit and we apply KVL to each loop individually Starting on the left side of the circuit V1 + 5 + 10 = 0 V1 = 5 + 10 = 15 V Next, moving to the loop on the top-right side of the circuit 5 + 7 + V2 + 3 = 0 V2 = 5 7 3 = 5 V Since V2 < 0, the actual polarity is the opposite of what is shown in the gure, ie, the actual +/ signs are reversed for V2 Continuing by considering the
Kirchhoff s Laws and Resistance
loop in the lower-right part of the circuit 10 3 + V3 = 0 V3 = 10 + 3 = 13 V Now we ve found all the voltages in the circuit, but we must note one other fact KVL is a general law that applies to the voltages around any loop KVL can therefore be applied to the outside loop in Fig 2-6 Hence V1 + 7 + V2 + V3 = 0 Is this true given the results we have found Let s insert the numbers V1 + 7 + V2 + V3 = 15 + 7 5 + 13 = 15 + 2 + 13 = 0 So KVL is indeed satis ed for the outer loop
The Resistor
In this chapter we study our rst circuit element in detail, the resistor As we will see, the resistor is actually a very simple device, so our analysis won t change too much at this point We will just have to do a bit of extra algebra The operation of a resistor is based on the following fact from physics As we know, a current is a ow of charges in other words, the charges in a material are moving in a given direction at some speed As the charges move through the material, they are going to collide with atoms that are xed in place in the form of a crystalline lattice As the charges move, they follow a process whereby they gain speed, move some distance, then collide with an atom, and have to start all over again To get them going we need to apply some kind of external force The external force is applied by impressing an electric eld on the material For small velocities the current density J, which is coulombs per cubic meter in SI units, is related to the electric eld via a linear relation of the form J = E (23)
In short, an applied voltage (and hence electric eld) gives the charges the energy they need to maintain their motion and keep the current going The constant of proportionality, which we have denoted by , is the conductance of the material The larger the is, the larger the current density J is for a given
Circuit Analysis Demysti ed
electric eld Metals such as copper or aluminum, which are good conductors, have very large values of , while a material like glass or wood will have a small value of The inverse of conductivity is resistivity, which is denoted by the Greek symbol Resistivity is the inverse of conductivity = So we could write (23) as E = J (24) 1
The units of resistivity are ohm-meters However, in circuit analysis where we concern ourselves with lumped elements we are more interested in resistance (a lumped element is one that has no spatial variation of v or i over the dimensions of the element) The dimensions of the element will not be important; only the global properties of the element are of concern to us We measure resistance in ohms, which are denoted by the upper case Greek character omega (ohms) (25)
Resistance is usually denoted by R, which is a constant of proportionality between voltage and current This relationship comes straight from (24) and is called Ohm s law after its discoverer In terms of voltage and current, it is written as follows V = RI (26)
It is possible for resistance to vary with time, but in many if not most cases it is a constant If the current and voltage vary with time, then Ohm s law can be written as v(t) = Ri(t) (27)
One way to think about what resistance does is to rewrite Ohm s law so that we have the current in terms of the voltage That is (26) can be written as I = V R
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