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Unformatted text preview: 121 CHAPTER 4 Exercises E4.1 The voltage across the circuit is given by Equation 4.8: ) / exp( ) ( RC t V t v i C − = in which V i is the initial voltage. At the time t 1% for which the voltage reaches 1% of the initial value, we have ) / exp( 01 . % 1 RC t − = Taking the natural logarithm of both sides of the equation, we obtain RC t / 605 . 4 ) 01 . ln( % 1 − = − = Solving and substituting values, we find t 1% = 4.605 RC = 23.03 ms. E4.2 The exponential transient shown in Figure 4.4 is given by ) / exp( ) ( τ t V V t v s s C − − = Taking the derivative with respect to time, we have ) / exp( ) ( τ τ t V dt t dv s C − = Evaluating at t = 0, we find that the initial slope is . / τ S V Because this matches the slope of the straight line shown in Figure 4.4, we have shown that a line tangent to the exponential transient at the origin reaches the final value in one time constant. E4.3 (a) In dc steady state, the capacitances act as open circuits and the inductances act as short circuits. Thus the steadystate (i.e., t approaching infinity) equivalent circuit is: From this circuit, we see that A. 2 = a i Then ohm’s law gives the voltage as V. 50 = = a a Ri v 122 (b) The dc steadystate equivalent circuit is: Here the two 10 Ω resistances are in parallel with an equivalent resistance of 1/(1/10 + 1/10) = 5 Ω . This equivalent resistance is in series with the 5 Ω resistance. Thus the equivalent resistance seen by the source is 10 Ω , and A. 2 10 / 20 1 = = i Using the current division principle, this current splits equally between the two 10 Ω resistances, so we have A. 1 3 2 = = i i E4.4 (a) ms 1 100 / 1 . / 2 = = = R L τ (b) Just before the switch opens, the circuit is in dc steady state with an inductor current of A. 5 . 1 / 1 = R V s This current continues to flow in the inductor immediately after the switch opens so we have A 5 . 1 ) ( = + i . This current must flow (upward) through R 2 so the initial value of the voltage is V. 150 ) ( ) ( 2 − = + − = + i R v (c) We see that the initial magnitude of v ( t ) is ten times larger than the source voltage. (d) The voltage is given by ) 1000 exp( 150 ) / exp( ) ( 1 t t R L V t v s − − = − − = τ τ Let us denote the time at which the voltage reaches half of its initial magnitude as t H . Then we have ) 1000 exp( 5 . H t − = Solving and substituting values we obtain ms 6931 . ) 2 ln( 10 ) 5 . ln( 10 3 3 = = − = − − H t 123 E4.5 First we write a KCL equation for . ≥ t ∫ = + + t dx x v L R t v 2 ) ( 1 ) ( Taking the derivative of each term of this equation with respect to time and multiplying each term by R , we obtain: ) ( ) ( = + t v L R dt t dv The solution to this equation is of the form: ) / exp( ) ( τ t K t v − = in which s 2 . / = = R L τ is the time constant and K is a constant that must be chosen to fit the initial conditions in the circuit. Since the initial ( t = 0+) inductor current is specified to be zero, the initial current in the...
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This note was uploaded on 11/20/2011 for the course ELE 220 taught by Professor Anas during the Spring '11 term at American Dubai.
 Spring '11
 Anas

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