Electronics - RL Circuits - Discussion

X = 2*3.14*20,000*18*10^-6 = 2.26 ohm

Total Impedance = sqrt(10^2 + 2.26^2) = 10.25 ohm
Voltage Drop across L2 = 27*(2.26)/10.25=6V

Please explain clearly.

Voltage drop across L2=Vs (Xl/Z).
Since in a parallel circuit voltage is same in all branches.

How total impedance is calculated, I can't get that line.

R2 = 10 Ohms.
XL2 = 2 * Pi * f * L2.
= 2 * pi * 20000 * 18 * 10^-6,
= 2.262 Ohms.

Now, Z = Pol(R2, XL2).
= Pol(10, 2.262),
= 10.253 Ohms.

Finally, Voltage Dropped Across L2 = Vs * (XL2 / Z).

= 5.956 V,

= 6 V Approx,

Long way to understand this, since voltage\current phase in R and L are 90 degree separated, assume VL2 in L2 and VR2 in R2. But total V = 27 and I is the same in that node due to series path.
i.e Total V is same in parallel; R=10ohm; XL2=2*3.14*20K*18uh =2.262
V^2 = (V(L2)^2 + V(R2)^2)
(V(L2)= I * XL2= 2.262I; V(R2) = I * R2 = 10I
27^2 = (2.262^2 *I^2 + 10^2 * I^2).
729 = 105.11 * I^2.
I = sqr(729/105.11) = 6.935.
So, V(L2) = I * XL2.
V(L2) = 2.633 * 2.262 = 5.956.

XL2 = 2 * 3.14 * 20,000 * 18 * 10^-6 = 2.26 ohm.
Z2 = Sq,rt (R^2 + XL^2).
Z2 = Sq,rt (10^2 + 2.26^2).
Z2 = 10.25 Ohm.

Voltage drop across L2 = Vs (XL2/Z2).
Voltage drop across L2 = 27 (2.26 /10.25).
Voltage drop across L2 = 5.95 V (nearly = 6 V).