To review material from Lecture 8as design problems. Problems 3.13, 3.24 and 3.6
ID: 2083981 • Letter: T
Question
To review material from Lecture 8as design problems. Problems 3.13, 3.24 and 3.63 in "Problem Sets" section in Blackboard, Chapter 3 (solutions available). Problem 3.25 in "Problem Sets" section in Blackboard, Chapter 3(solution available) Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1-ms duration to the heart, which can be as a 500-ohm resistance. The peak amplitude of the pulses is required to be 5V.However, the battery delivers only 2.5-V Therefore, we decide to charge two equal-value capacitors in parallel from the 2.5-V battery and then switch the capacitors in series with the heart during the 1-ms pulse. What is the minimum value of the capacitances required so the output pulse amplitude remains between 4.9 V and 5.0 V throughout the 1-ms duration? If the pulses occur once every second, what is the average current drain from the battery? Use approximate calculations, assuming constant current during the output pulse. Find the ampere-hour rating of the battery so it lasts for five years. Jenny the Circuit Builder has seven components on her workbench. She is trying to build a circuit that stores the maximum amount of energy possible. Her only energy source is a battery with a model as shown below (voltage source at 9V and internal resistance of 2 ohm). Real inductors and capacitors also have some internal resistance associated with them. The six components have effective series resistance elements as shown below. [These are values from real capacitors and inductors that you can buy in the exist1-exist5 range]. (a) Find which single component will hold the most energy in steady state while connected across the terminals of the battery. (b) Using as many components as you like, design a circuit that can hold the most energy possible. Explain what would happen is your circuit was disconnected from the battery. Try simulating it.Explanation / Answer
Solution:
A) The energy through stored in capacitor and inductor is given by the formula below
Ec = (1/2) x C x Vc2 where Vc is the voltage across capacitor
EL = (1/2) x L x I2 where I is the inductor current
From above expressions, it is clear that capacitor or inductor holds maximum energy if it's size is more.
There are two possibilities, the component having capacitance 10mF or inductance 47mH can be connected.
among them we need to think about the impedance provided by each of them and it is clear that capacitance provides less impedance for the same value of w then inductive impedance.
Hence, The component of having capacitance 10 mF connected across the terminals of the battery.
B) The combination of components of having capacitance 10mF and inductance 47 mH will be connected in series to hold the maximum energy with less loss across the resistor values because their values are less in both the components and the reason behind this could be same as part (A).
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