Answer to Question #200145 in Electrical Engineering for Rownak Jahan

Question

Slove this question according to figure 04.

1. Find the Thevenin equivalent circuit looking from the terminals a-b as shown in Figure 04.
2. In Figure 04, assume that the load is purely resistive (RL), determine the value of the load resistor RL across terminals a-b for maximum average-power transfer, and calculate the RMS value of the load current when RL is connected between terminals a-b.
3.  Calculate the complex power drawn by the load when RL is connected between terminals a-b.

Answer 1

(a) Let current I flow in $3\Omega$ resistor as no current will flow in $\perp\Omega$ resistor due to open.

By KCL at node b

Current in S $\Omega$ resistor $=(I+2)A$ ,so By KCL in loop

$24=3I+4I+5(I+2)+2I$

$24=14I+10$

$I=\perp A$

So, $V_{th}=(5\Omega)(I+2)A$

$=5*(\perp+2)$

$=15V$

(b) First we have to find Rth (thevenin equivalent seen from b-c node) for this short voltage source and open current source.

As $\perp\Omega$ is open so not take part

$\therefore R_{th} =(2\Omega+3\Omega+4\Omega)//(sn)$

$=(g\Omega)//(5\Omega)$

$R_{th}=\frac{g*5}{g+5}=\frac{45}{15}\Omega=3.214\Omega$

So, therenin equivalent current for maximum power transfer

$R_L=R_{th}$

$\therefore R_l=3.214\Omega$

(c) As $R_l=R_{th}=3.214$ for maximum power

So, maximum power transfer to $R_l=I_{max}^2*R_l$

$=(\frac{V_{th}}{R{th}+R_1})^2*R_l$

$=(\frac{15}{3.214+3.214})*3.214$

$=17.5 watts$