Electrical Engineering - Circuit Theorems and Conversions - Discussion

Anyone give me an explanation for this one.

You can solve this by assigning three loop currents into the 3 visible loops and solving the resultant equations for the currents. Then the voltage through the load resistor can be found.

In a simple way use R = V/I and validate with options as 4.5/4.5*10^-3 =1K. So option C is correct.

Right @Ashu.

Explain it clearly.

Nice one @Ashu.

Thank you all.

For the Thevenin equivalent circuit for the network external to branch.

From the Thevenin equivalent circuit, Vth and Rth then the RL, use a voltage divider to get the VL.

Then convert the Thevenin to Norton's, IN and RN then RL, use a current divider to get IL.

To get the Rth.
Rth = ((R1xR2)/R1+R2) + ((R3xR4)/R3+R4).
Rth = 422 approx.

Use voltage divider of R1 and R2. use voltage divider of R3 and R4. then subtract voltage divider of R1 and R2 to voltage divider of R3 and R4.

=> V12 = Vs( R2/(R2 + R1)).
V12 = 22.6667V.

=> V34 = Vs( R4/(R4 + R3)).
V34 = 16.31 V.

To get the Vth.
Vth = V12 - V34.
Then, Vth = 6.36 V.

To get the Voltage load, use a voltage divider of Vth, Rth and RL.
VL = 6.36( 1000/(1000+422)).
VL = 4.5 V.

To get the current load, use the current divider.
Vth = Ith(Rth).
Here; Ith = 0.01517 A.

So, IL = Ith(Rth/(Rth +RL)).
= 0.01517 (422/(422 + 1000).
Therefore IL = 4.5mA.