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Part I
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Circuits
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Figure 54 Decaying and rising exponential responses
SOLUTION OF CIRCUITS CONTAINING DYNAMIC ELEMENTS
The major difference between the analysis of the resistive circuits studied in s 2 and 3 and the circuits we will explore in the remainder of this chapter is that now the equations that result from applying Kirchhoff s laws are differential equations, as opposed to the algebraic equations obtained in solving resistive circuits Consider, for example, the circuit of Figure 55, which consists of the series connection of a voltage source, a resistor, and a capacitor Applying KVL around the loop, we may obtain the following equation: vS (t) vR (t) vC (t) = 0 (51)
A circuit containing energy-storage elements is described by a differential equation The differential equation describing the series RC circuit shown is diC dv 1 + i = S dt RC C dt + vR _ R
+ vS (t) _
Observing that iR = iC , we may combine equation 51 with the de ning equation for the capacitor (equation 46) to obtain vS (t) RiC (t) 1 C
t
iC C + vC (t) _
iC dt = 0
(52)
Equation 52 is an integral equation, which may be converted to the more familiar form of a differential equation by differentiating both sides of the equation, and recalling that d dt
t
iC (t ) dt
= iC (t)
(53)
Figure 55 Circuit containing energy-storage element
to obtain the following differential equation: diC 1 dvS 1 + iC = dt RC R dt where the argument (t) has been dropped for ease of notation (54)
5
Transient Analysis
Observe that in equation 54, the independent variable is the series current owing in the circuit, and that this is not the only equation that describes the series RC circuit If, instead of applying KVL, for example, we had applied KCL at the node connecting the resistor to the capacitor, we would have obtained the following relationship: iR = or dvC 1 1 + vC = vS dt RC RC (56) vS vC dvC = iC = C R dt (55)
Note the similarity between equations 54 and 56 The left-hand side of both equations is identical, except for the variable, while the right-hand side takes a slightly different form The solution of either equation is suf cient, however, to determine all voltages and currents in the circuit The following example illustrates the derivation of the differential equation for another simple circuit containing an energy-storage element
EXAMPLE 51 Writing the Differential Equation of an RL Circuit
Problem
+ vR _ R1 vS (t) + _ iR1 iL L + vL _ iR2 R2
Derive the differential equation of the circuit shown in Figure 56
Solution
Known Quantities: R1 = 10
; R2 = 5
; L = 04 H
Find: The differential equation in iL (t) Figure 56 Assumptions: None Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation Note that the top node voltage is the inductor voltage, vL
iR1 iL iR2 = 0 vL vS vL iL =0 R1 R2 Next, use the de nition of inductor voltage to eliminate the variable vL from the nodal equation L diL L diL vS iL =0 R1 R1 dt R2 dt R2 R 1 R2 diL iL = vS + dt L (R1 + R2 ) L (R1 + R2 ) Substituting numerical values, we obtain the following differential equation: diL + 833iL = 0833vS dt
Part I
Circuits
Comments: Deriving differential equations for dynamic circuits requires the same basic circuit analysis skills that were developed in 3 The only difference is the introduction of integral or derivative terms originating from the de ning relations for capacitors and inductors
We can generalize the results presented in the preceding pages by observing that any circuit containing a single energy-storage element can be described by a differential equation of the form a1 dx(t) + a0 x(t) = f (t) dx (57)
where x(t) represents the capacitor voltage in the circuit of Figure 55 and the inductor current in the circuit of Figure 56, and where the constants a0 and a1 consist of combinations of circuit element parameters Equation 57 is a rstorder ordinary differential equation with constant coef cients The equation is said to be of rst order because the highest derivative present is of rst order; it is said to be ordinary because the derivative that appears in it is an ordinary derivative (in contrast to a partial derivative); and the coef cients of the differential equation are constant in that they depend only on the values of resistors, capacitors, or inductors in the circuit, and not, for example, on time, voltage, or current Consider now a circuit that contains two energy-storage elements, such as that shown in Figure 57 Application of KVL results in the following equation: Ri(t) L 1 di(t) dt C
t
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