Electrical Engineering - Circuit Theorems and Conversions - Discussion

Rth = R1||R2 + R3||R4 = 422.
Vth = Va-Vb.
Va = R2/(R1+R2)*Vs = 22.666.
Vb = R4/(R3+R4)*Vs = 16.31.
Vth = 6.356 V.

What about vth then?

Vth = {30*(470/470+560)}-{30*(220/220+680)}.
Vth = 6.3V.

Rth = (470+220)//(680+560).
Rth = 443 ohm.

So, option 1 is right.

Rth = (R1 // R2) + ( R3 // R4).
= 220||680 + 470||560 = 421.75 (422).

Vth is the open circuited voltage therefore find the effective resistance neglecting RL
R eff = R1 + R2 II R3 + R4.
= 220 + 680 II 560 + 470 = 900 II 1030= (900 * 1030) /(900 + 1030) = 480.31.

Now find the total current.
I = V/Reff =30/480.31 = 0.062.

Find current through each parallel branch using current division:
I1 = (0.062 * 1030)/(1030 + 900) = 0.033.
I2 = 0.062 - 0.033 = 0.029.

Vth= VA - VB = V470 - V220.
=(470 * 0.029) - (220 * 0.033) = 6.37 v.

Thanks for explaining it.

From the circuit

Vth= ((R1||R3)*Vs)/(R1+R3);

(note: R1||R3 means R1 in parllel with R3, R1||R3=149.8);

on sloving we get

Vth=(149.8*30)/690

Vth=6.5 (app. equal to 6.4v;

from the circuit we get

Rth=(R1+R3)||(R2+R4)

Rth=690||1240;

Rth=443.31(app. equal to 422 ohm)

so option A

Thevenin's resistance is (R1 // R2) + ( R3 // R4).

Thevenin's voltage is obtained by applying potential divider rule,

Vth = Voltage across R2 - voltage across R4.

R1 parallel to R2& R3 is parallel to R4 and the result of R1||R2&R3||R4 are in series.

= 220||680+470||560 = 421.75 (422).

I agree with answers. But I am not getting series and parallel combination. Please tell me if you know.

Well said, Thank you all.