1.11 Solutions to Exercises

Dear Reader:

The remaining pages on this chapter contain the solutions to the exercises.

You must, for your benefit, make an honest effort to find the solutions to the exercises without first looking at the solutions that follow. It is recommended that first you go through and work out those you feel that you know. For the exercises that you are uncertain, review this chapter and try again. Refer to the solutions as a last resort and rework those exercises at a later date.

You should follow this practice with the rest of the exercises of this book.

1. 
   
   

   and since , the above becomes

   
   

   From the characteristic equation

   
   

   we get s ₁ = s ₂ = ? 25 (critical damping) and ? _S = R/2L = 25

   The total solution is

   
   

   With the first initial condition v _C( 0 ^?) = 0 the above expression becomes 0 = 100 + e ⁰ ( k ₁ + 0) or k ₁ = ?100 and by substitution into (1) we get

   
   

   To evaluate k ₂ we make use of the second initial condition i _L( 0 ^?) = 0 and since i _L = i _C, and i = i _C = C( dv _C)/( dt), we differentiate (2) using the following MATLAB code:

   <span class="serif">syms t k2v0=100+exp(-25*t)*(k2*t-100); v1=diff(v0)</span>v1 =-25*exp(-25*t)*(k2*t-100)+exp(-25*t)*k2

   Thus,

   
   

   and

   
   

   Also,