Electrical Engineering - Circuit Theorems in AC Analysis - Discussion

It is very simple.
1) As we know voltage across parallel is same so XL is useless to calculate;
2) To find vth RL is open circuit so remove RL.
3) Now we have VS= 6 V and Xc= 12 K is in series with R = 15 K.
4) Z = (R^2+Xc^2)^1/2 = 19.2 K.
5) Vth = R/Z * Vs = .78*6 = 4.69.
6) θ = tan-1 ( Xc/R) = .6747 rad * 57 = 38.5°.

It's very good explanation @Nhuss.

The Vth across RL is equal to the voltage across the resistor R. So for determining the voltage across R we need to find out the current flowing through R. The Current across R would be equal to I (across R) = VS/(R+XC)

So the Voltage across R would be equal to IR.

Vth = I (across R) * R.
= (VS*R)/(R + XC).

Put the values of the respective parameters and the answers would be option b = 4.69<38.7° V.

Eth is the voltage across R resistance.

So, Eth = 6*15/(15 - 12j) = 4.68 < 38.7.

If you write equation for loop 1(branch with the inductor) and loop 2(the branch with the capacitor and R after removing RL):

Loop 1: I1Xl = 6V
Loop 2: I2(15000-j12000) = 6 (since we are interested Vth across R, we can find I2, then find the voltage across R. Because Rl is in parallel with R, the voltage across R is the same as Vth)

I2 = 6V/(15000-j120000)= 3.1235*10^-4<theta = 38.66>

Vr=Vth = 15000 * 3.1235*10^-4<theta = 38.66> = 4.685<theta = 38.66>

Hope this helps!

XL = 12 K OHMS.
XC = 12 K OHMS.
RL = 49 K OHMS.
R = 15 K OHMS.
VS = 6 < 0 V.

Z = SQRT ((R+RL)^2 + (XL-XC)^2).

Z = R + RL = 49 + 15 = 64 K OHMS.

I = V/Z = 6/64K = 0.09375 mA.

VTH = 0.09375 mA x 49 k OHMS = 4.59 V.

Use kcl to find current in 15 kohm resistance. Then find voltage and that will answer.

Can anybody tell true procedure?

Can you please elaborate the explanation by substituting values @Nikhil Jain.

The value of C is j12 kohm and l is j12 kohm.

First remove RL and consider vth at the place of Rl then apply nodal analysis. You'll get the answer