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Determine the Norton equivalent network seen by the capacitor in Figure P4.71. Use the result and current division to find
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Principles and Applications of Electrical Engineering
- Due to components not shown in the figure, the circuit of Figure P4.41 has i L ( 0 )= I i . a. Write an expression for i L (t) for t≥0. b. Find an expression for the power delivered to the resistance as a function of time. c. Integrate the power delivered to the resistance from t=0 to t=∞, and show that the result is equal to the initial energy stored in the inductancearrow_forwardThe capacitor model we have used so far has beentreated as an ideal circuit element. A more accuratemodel for a capacitor is shown in Figure P4.67. Theideal capacitor, C, has a large “leakage” resistance, RC,in parallel with it. RC models the leakage currentthrough the capacitor. R1 and R2 represent the leadwire resistances, and L1 and L2 represent the lead wireinductances.a. If C = 1 μF, RC = 100 MΩ, R1 = R2 = 1 μΩ andL1 = L2 = 0.1 μH, find the equivalent impedanceseen at the terminals a and b as a function offrequency ω.b. Find the range of frequencies for which Zab iscapacitive, i.e., Xab > 10|Rab.Hint: Assume that RC is is much greater than 1/wC so thatyou can replace RC by an infinite resistance in part b.arrow_forwardThe circuit shown in Figure P4.26 is operating in steady state. Determine the values of i L,v x ,and v C .arrow_forward
- We know that the capacitor shown in Figure P4.11 is charged to a voltage of 10 V priorto t=0.a. Find expressions for the voltage across the capacitor vC(t) and the voltage across theresistor vR(t) for all time.b. Find an expression for the power delivered to the resistor.c. Integrate the power from t=0 to t=∞ to find the energy delivered.d. Show that the energy delivered to the resistor is equal to the energy stored in thecapacitor prior to t=0.arrow_forwardDetermine expressions for and sketch i s ( t ) to scale versus time for −0.2≤t≤1.0 s for the circuit of Figure P4.37.arrow_forwardThe initial voltage across the capacitor shown in Figure P4.3 is v C ( 0+ )=−10 V. Find anexpression for the voltage across the capacitor as a function of time. Also, determine the time t0at which the voltage crosses zero.arrow_forward
- Consider the circuit shown in Figure P4.55. a. Write the differential equation for v(t).b. Find the time constant and the form of the complementary solution.c. Usually, for an exponential forcing function like this, we would try a particular solution ofthe form vp(t) = K exp (−10t). Why doesn’t that work in this case?d. Find the particular solution. [Hint: Try a particular solution of the form vp(t)=K t exp (−10t). How ]e. Find the complete solution for v(t).arrow_forwardAt t=0 a charged 10{μF capacitance is connected to a voltmeter, as shown in Figure P4.5. The meter can be modeled as a resistance. At t=0 the meter reads 50 V. At t=30s, the reading is 25 V. Find the resistance of the voltmeter.arrow_forwardConsider the circuit shown in Figure P4.50. The initial current in the inductor is i s ( 0+)=0. Write the differential equation for i s(t) and solve. [Hint: Try a particular solution of the form i sp ( t )=A cos( 300t )+B sin( 300t ).]arrow_forward
- Consider the circuit shown in Figure P4.54. a. Write the differential equation for i(t). b. Find the time constant and the form of the complementary solution. c. Usually, for an exponential forcing function like this, we would try a particular solution of the form ip(t)=K exp (−3t). Why doesn’t that work in this case? d. Find the particular solution. [Hint: Try a particular solution of the form ip(t)=K t exp(−3t).] e. Find the complete solution for i(t).arrow_forwardSolve for i L ( t ) for t>0 in the circuit of Figure P4.48. You will need to make an educated guess as to the form of the particular solution. [Hint: The particular solution includes terms with the same functional forms as the terms found in the forcing function and its derivatives.]arrow_forwardThe voltage across an inductor plotted as a functionof time is shown in Figure P4.14. If L = 0.75 mH,determine the current through the inductor att = 15 μs.arrow_forward
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