Question

The state equation for the current $I_1$ shown in the network shown below in terms of the voltage $V_x$ and the independent source V, is given by

• A. $\frac{d{I}_1}{dt}=-1.4V_x-3.75I_1+\frac{5}{4}V$
• B. $\frac{d{I}_1}{dt}=-1.4V_x-3.75I_1-\frac{5}{4}V$
• C. $\frac{d{I}_1}{dt}=-1.4V_x+3.75I_1+\frac{5}{4}V$
• D. $\frac{d{I}_1}{dt}=-1.4V_x+3.75I_1-\frac{5}{4}V$
  

Answer 1

The correct option is A $\frac{d{I}_1}{dt}=-1.4V_x-3.75I_1+\frac{5}{4}V$

Any state equation represents the dynamical behaviour of the given network. State equations usually follow a specific 'format' while being represented. On the left side od each state equation, the derivative of only one variable is used. On the right hand side a mathematical function is represented involving any or all the state variable and the sources.


Using KVL in Loop-I,

$V-3(I_1+I_2)-V_x-0.5\frac{d{I}_1}{dt}=0$

Using KVL in loop-II,

$0.5\frac{d{I}_1}{dt}-5I_2+0.2V_x=0$

Eliminating $I_2$ from equation (i) and (ii), we get,

$\frac{d{I}_1}{dt}=-1.4V_x-3.75I_1+\frac{5}{4}V$