Question

1). Calculate the current intensity \( I_x \)

2). Générateur de Thévenin entre les bornes 𝑎−𝑏

Answer 1

Solution

1). Calculate the current intensity \( I_x \)

Redraw the circuit for the ;opp analysis, 

Apply KCL at  node A.

\( I_1+I_x =1.5I_x \)

\( I_1=0.5I_x \)   (1)

Redraw the circuit for the ;opp analysis, 

Apply KVL in both mesh.

\( -6-5I_1+3I_x+4I_x=0 \)

From equation (1) :

\( -6-5(0.5I_x)+3I_x+4I_x=0 \)

\( -6-2.5I_x+3I_x+4I_x=0 \)

\( -6+4.5I_x=0 \)

\( 4.5I_x=6=>I_x=\frac{6}{4.5}=1.33A \)

2). Générateur de Thévenin entre les bornes 𝑎−𝑏

For \( {E}_{th} \)

\( {E}_{th}=4I_x \)

       \( =4(\frac{6}{4.5})=5.33V \)

To calculate \( {I}_{sc} \) , short the terminal a-b , The circuit would be drawn as,

Apply KVL in large loop .

\( -6-5I_1+3{I}_{sc}=0 \)

Form equation (1) :

\( -6-5(0.5I_x)+3{I}_{sc}=0 \)

\( -6-2.5{I}_{sc}+3{I}_{sc}=0 \)

\( 0.5{I}_{sc}=6=>{I}_{sc}=\frac{6}{0.5}=12A \)

Now, calculate \( {R}_{th} \)

\( {R}_{th}=\frac{{E}_{th}}{{I}_{sc}}=\frac{5.33}{12}=0.44\Omega \)

1). Thus, \( I_x=1.33A \)

2). Thus, \( {E}_{th}=5.33V \) & \( {R}_{th}=0.44\Omega \)