Question

In the circuit shown in figure below,

the maximum average power absorbed by the load $R_L$ is

• A. 29.39 W
• B. 39.29 W
• C. 28.40 W
• D. 42.26 W

Answer 1

The correct option is B 39.29 W

$Z_{th}=(40-j30)\parallel j20$
$=\frac{j20(40-j30)}{j20+40-j30}=(9.412+j22.35)\Omega$

By voltage division,

$V_{th}=\frac{j20}{j20+40-j30}\left(150\angle {30}^o\right)$

$=72.76\angle {134}^{o}V$

The value of $R_L$ that will absorb the maximum average power is

$R_L=|Z_{th}|=\sqrt{(9.412{)}^2+(22.35{)}^2}=24.25\Omega$

The current through the load is,

$I=\frac{V_{th}}{Z_{th}+R_L}=\frac{72.76\angle {134}^o}{33.66+j22.35}=1.8\angle {100.416}^o$

The maximum average power absorbed by $R_L$ is

$P_{max}=\frac{1}{2}|I{|}^2{R}_L=\frac{1}{2}\left(1.8{)}^2\right(24.25)$
=$39.285=39.29$ W