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Satwardhan N.
Undergrad at BITS Pilani
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Electrical Engineering
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Question:

A capacitor is charged such that it develops a potential difference of 5 V across its plates.At t=0 it is connected to an inductor forming an LC circuit. What will be the maximum value( magnitude) of the current flowing through this circuit for time t>0? Inductor value(L)=2 H.Capacitor value(C)=2 F.

Satwardhan N.
Answer:

This problem deals with the concept of energy balance. Total energy at t<0 in the capacitor : $$0.5CV^2=25 J$$(feeding in the approprate values) Total energy dissipated through the inductor:$$0.5LI^2$$ Electrical energy of the capacitor will get converted into energy of the inductor,which will again get converted to electrical energy of capacitor and this cycle will go on. $$0.5CV^2=0.5LI^2$$ $$0.5LI^2=25 J$$ $$I=5 A$$

C++ Programming
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Question:

Print the numbers from 1-100 in serial order, each on a newline

Satwardhan N.
Answer:

#include <iostream.h> #include<conio.h> void main() { for (int i=0;i<100;i++) { cout<<i<<endl; } }

Differential Equations
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Question:

Obtain the solution for the equation : $$4dy/dt$$ + 3$$y$$ = $$y^2$$

Satwardhan N.
Answer:

$$4dy/dt$$ = $$y^2$$- $$3y$$ $$4dy/(y^2-3y) = dt$$ $$dy/y(y-3) =dt/4$$ Now , by seperating the term on the LHS into partial fractions we get: $$(-dy/3)( (1/y) - (1/(y-3)))=dt/4$$ $$dy/y - dy/(y-3)=-3dt/4$$ Integrating both sides we get: $$lny-ln(y-3) = -3t/4 + c$$ (c is the integration constant) $$ln(y/(y-3))=-3t/4 +c$$ $$y/(y-3)=e^(-3t/4+c)$$ let $$e^c=k$$(some constant) hence, $$y/(y-3)=ke^-3t/4$$ by cross multiplying we get, $$k(y-3)/y=e^(3t/4)$$ $$1-(3/y)=(e^(3t/4))/k$$ $$y=3k/(k-(e^(-3t/4)))$$ Boom , there goes the solution!

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