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# Tutor profile: Satwardhan N.

Inactive
Satwardhan N.
Undergrad at BITS Pilani
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## Questions

### Subject: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.

Inactive
Satwardhan N.

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\$\$

### Subject:C++ Programming

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Question:

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

Inactive
Satwardhan N.

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

### Subject:Differential Equations

TutorMe
Question:

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

Inactive
Satwardhan N.

\$\$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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