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# Answer to Question #200145 in Electrical Engineering for Rownak Jahan

Question #200145

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.
1
2021-05-31T06:20:02-0400

(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"

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