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Creator Code 128 Code Set C in VS .NET Fig 3-39 Circuit for Problem 1

Fig 3-39 Circuit for Problem 1
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Thevenin s and Norton s Theorems
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Fig 3-40 Network for Problem 3
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2 What is the equivalent resistance if a 6 resistor is in parallel with a 4 resistor 3 Find the equivalent resistance for the circuit in Fig 3-40 4 Find the Thevenin equivalent voltage and resistance for the circuit shown in Fig 3-41 Do this using Thevenin s theorem, and then show that you get the same answer using the Karni method 5 Find the current i (t) in Fig 3-41
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i4 4 6A 2e t + i(t) 01i3
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Fig 3-41 Circuit for Problems 4 and 5 Borrowed from a circuit analysis exam given by Shlomo Karni in 1990
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We have already seen two important network theorems Thevenin s and Norton s theorems In this chapter we will introduce other theorems that can be used to simplify network analysis
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Superposition
Consider a circuit that contains multiple voltage and current sources If all of the elements in the circuit are linear, we can simplify analysis by considering the effect of each source individually and then adding up the results How does this work A given component will have a voltage across it and a current through it due to each source We calculate those voltages and currents considering each source individually Then we add up all the currents and all the voltages to get the total current and voltage for the component This is the essence of the superposition theorem Let s quantify this Let a circuit contain a set of n sources where the ith source is denoted si Now suppose that the response of the circuit to si alone is ai si , where ai is a constant Then the total response
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Fig 4-1 We can apply the superposition theorem to this linear circuit, because it contains two sources
of the circuit to all of the sources is found by adding up the individual responses r = a1 s1 + a2 s2 + + an sn (41)
Let s see how to apply the superposition theorem to the circuit shown in Fig 4-1 EXAMPLE 4-1 Use the superposition theorem to nd the current through the 4 Fig 4-1 resistor in
SOLUTION First let s solve the circuit by using ordinary techniques We begin by applying KVL to the left pane in Fig 4-1 We nd 5 + 4I1 + 5(I1 I2 ) = 0 Collecting and rearranging terms gives 9I1 5I2 = 5 Now we apply KVL to the right pane in Fig 4-1 We have 3 + 2I2 + 5(I2 I1 ) = 0 Collecting terms we obtain 5I1 + 7I2 = 3
Circuit Analysis Demysti ed
To nd the two unknown currents we arrange the terms as the following system of equations 9 5 5 5 7 3
The matrix on the left contains the coef cients of each current (resistances) while the column vector on the right is due to the sources We use Cramer s rule to determine the value of the two unknown currents For the rst current we have 5 5 3 7 9 5 5 7 20 35 15 = A 63 25 38
I1 =
This is the unknown current through the 4 resistor that we will nd using superposition in a moment The other current is found to be 9 5 9 5 5 3 5 7 8 27 + 25 = A 63 25 38
I2 =
Now let s solve for the rst current using superposition We will do this by replacing each voltage source by a short circuit in turn We begin by leaving the 5 V source intact and setting the 3 V source to zero This results in the circuit shown in Fig 4-2 We will illustrate the method by going through the same process of applying KVL In the rst pane, the equation is 5 + 4I1 + 5(I1 I2 ) = 0
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