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Qaisersatti
 2 years ago
Best ResponseYou've already chosen the best response.1Let's look at the effect of the inductance of a wire lead on a component. A typical 1/4 Watt discrete resistor comes with 1.5 inch leads of AWG #20 wire, one for each terminal. We usually mount such a part using only a bit of that wire. Suppose we end up with a total of 2cm of wire in series with that resistor. AWG #20 wire has a diameter of 0.812mm. So 2cm of this wire has an inductance of about 15.44nH. (The estimate of inductance is computed from from a magic formula that is not part of this subject. You can learn about this in an electromagnetic theory subject.) We will ignore the resistance of the 2cm of wire... Suppose we have a voltage source that turns on (it goes from 0 to V) at t=0, and suppose it is connected to an RO load through a resistor RS and a wire with inductance L. Let's model the situation as follows: Assume that the initial voltage across the load is zero. What is the final voltage across the load? Write an algebraic expression for this final value in terms of the parameters given: Write an algebraic expression for the time constant of this circuit in terms of the parameters given: Now, let's get down to numbers. Let V=3.3V, RS=22.0Ω, RO=50.0Ω, and L=15.44nH. How much time, in nanoseconds, does it take for the output voltage to reach vO=1.65V? This is a pretty significant amount of time! For comparison, how long, in nanoseconds, does light take to go 2cm? Remember, the speed of light is 299792458m/s.

Qaisersatti
 2 years ago
Best ResponseYou've already chosen the best response.1has been trying whole day....

Nurali
 2 years ago
Best ResponseYou've already chosen the best response.6(V*RO)/(RO+RS), L/(RO + RS), 0.273, 0.0667

monishvster
 one year ago
Best ResponseYou've already chosen the best response.0@nurali How you got 3rd part??
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