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Fig 6-14 A plot of the current found in Example 6-10
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The plot, shown in Fig 6-14, shows the current approaching the nal value of 2 A in about 125 s, as would be expected from the time constant EXAMPLE 6-11 A resistor R = 12 and inductor L = 4 H are connected in series with a sinusoidal voltage source v s (t) = 20 sin 30t Find the total response of the circuit for i(t) The initial current is zero SOLUTION The differential equation describing the circuit is 4 di + 12i(t) = 20 sin 30t dt
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The zero-input response is found by setting the value due to the source (the input on the right-hand side of the equation) to zero This gives di H + 3i H (t) = 0 dt We nd the solution to be i H (t) = K e 3t where K is a constant to be determined from the initial condition Now we need to solve for the zero-state response, or the particular solution for the current
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The inhomogeneous term is
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20 sin 30t Suggesting a solution of the form i p (t) = A sin(30t + ) The derivative of this expression is di p = 30A cos(30t + ) dt Now use the following trig identities sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y sin x sin y To write i p (t) = A sin(30t + ) = A sin 30t cos + A cos 30t sin di p = 30A cos(30t + ) = 30A cos 30t cos 30A sin 30t sin dt Then we have di p + 3i p (t) = 30A cos 30t cos 30A sin 30t sin dt + 3A sin 30t cos + 3A cos 30t sin Grouping terms and setting equal to the inhomogeneous term due to the voltage source we have A cos 30t(30 cos + 3 sin ) + A sin 30t(3 cos 30 sin ) = 5 sin 30t Since there is no term involving cos 30t on the right-hand side, it must be true that 30 cos + 3 sin = 0
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This leads to the relation tan = 10 = 84 Now, 3 cos( 84 ) 30 sin ( 84 ) = 3(0105) 30( 0995) = 30155 So we can solve for A Given that 30 cos + 3 sin = 0, we are left with A sin 30t(3 cos 30 sin ) = 5 sin 30t 5 A= = 0166 3 cos 30 sin So the particular solution, which represents the zero-state solution, is i p (t) = 0166 sin(30t 84 ) The total solution is i(t) = i p (t) + i H (t) = 0166 sin(30t 84 ) + K e 3t Setting this equal to zero gives K = 0166 sin( 84 ) = +0166 The total solution is therefore i(t) = 0166(sin(30t 84 ) + e 3t )
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Second-Order Circuits
A second-order circuit is one that includes capacitors and inductors in a single circuit Such a circuit is called second order because of the nature of the currentvoltage relations for inductors and capacitors When the analysis is done, a second-order differential equation for the current or the voltage will result Second-order circuits are subject to a phenomenon known as damping Before looking speci cally at electric circuits, consider an arbitrary second-order differential equation of the form:
2 s 2 + 2 n s + n = 0
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where n is the undamped natural frequency and is the damping ratio The dynamic behavior of the system is then described in terms of these two parameters and n To determine the behavior of a system, we look at the damping ratio There are three possibilities: If 0 < < 1, the roots are complex conjugates and the system is underdamped and oscillatory If = 1, the roots of the system are equal and the response is critically damped If > 1, the roots are negative, real, and unequal The system response is overdamped With this in mind, consider the RLC circuit second-order equation (628) that is speci c for a resistor, inductor, and capacitor in series: s2 +