iC dt

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(426)

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Figure 425 Circuit containing energy-storage element

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Equation 426 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

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t

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iC (t ) dt

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= iC (t)

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(427)

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AC Network Analysis

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to obtain the following differential equation: diC 1 1 dvS + iC = dt RC R dt (428)

where the argument (t) has been dropped for ease of notation Observe that in equation 428, 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 (430) vS vC dvC = iC = C R dt (429)

Note the similarity between equations 428 and 430 The left-hand side of both equations is identical, except for the independent 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 Forced Response of Circuits Excited by Sinusoidal Sources Consider again the circuit of Figure 425, where now the external source produces a sinusoidal voltage, described by the expression vS (t) = V cos( t) (431)

Substituting the expression V cos( t) in place of the source voltage, vS (t), in the differential equation obtained earlier (equation 430), we obtain the following differential equation: d 1 1 vC + vC = V cos t dt RC RC (432)

Since the forcing function is a sinusoid, the solution may also be assumed to be of the same form An expression for vC (t) is then the following: vC (t) = A sin t + B cos t which is equivalent to vC (t) = C cos( t + ) (434) (433)

Substituting equation 433 in the differential equation for vC (t) and solving for the coef cients A and B yields the expression A cos t B sin t + = 1 V cos t RC 1 (A sin t + B cos t) RC

(435)

Part I

Circuits

and if the coef cients of like terms are grouped, the following equation is obtained: B V A B sin t + A + RC RC RC cos t = 0 (436)

The coef cients of sin t and cos t must both be identically zero in order for equation 436 to hold Thus, A B = 0 RC and A + B V =0 RC RC (437)

The unknown coef cients, A and B, may now be determined by solving equation 437, to obtain: A= B= V RC 1 + 2 (RC)2 V 1 + 2 (RC)2 V RC V sin t + cos t 2 (RC)2 1+ 1 + 2 (RC)2

(438)

Thus, the solution for vC (t) may be written as follows: vC (t) = (439)

v (t) (V) vS (t) vC (t) 0 167 333 5 Time (ms)

This response is plotted in Figure 426 The solution method outlined in the previous paragraphs can become quite complicated for circuits containing a large number of elements; in particular, one may need to solve higher-order differential equations if more than one energystorage element is present in the circuit A simpler and preferred method for the solution of AC circuits will be presented in the next section This brief section has provided a simple, but complete, illustration of the key elements of AC circuit analysis These can be summarized in the following statement: