Electronics - RLC Circuits and Resonance - Discussion

Using reverse engineering method:

Since the circuit is in series connection.
V= √(Vr squared+(Vl-Vc)squared)
317=√(156 squared+(441-165) squared)
317=317.

Therefore the correct answer is A.

Good @Rajan. Thanks for explaining this.

@Mukul you are correct.

If you need to understand the solution check below.

Xl=(2.pi.f.L) => 2*3.14*1500*3 => 28,274ohm.
Xc= ( 1/2.pi.f.C) => 1/2*3.14*1500*0.01uF => 10,610ohm.

Vt=317v ( total).
Vr=I*R ( resis).
Vl=I*Xl (Induc).
Vc=I*Xc (caps).

(Note Vl and Vc are in reverse angle). So Vl - Vc in one vertical plane. Vr is horizontal.

I is same in all the devices but V is not same in degree so use Pythagoras, then we get,

sqr(Vt) = sqr(Vr) + sqr(Vl - Vc).
sqr(317) = sqr(I*R) + sqr(I*Xl - I*Xc).
sqr(317) = sqr(I) *( sqr(10K) + sqr(28,274 - 10,610) ).

sqr(I)= sqr(317)/ sqr(10K) + sqr(17,664).

I = sqrt(2.43895- e4).
I = 0.015617A.
= 15.617mA.

Vr = I*R = 0.015617*10k = 156V.
Vl = I*Xl = 0.015617 * 28,274 = 441V.
Vc = I*XC = 0.15617 * 10,610 = 165V.

Good, we need the clear basics concepts.

find X(l)=2*3.14*f*l kohm
then X(c)=1/3.14*f*c kohm
then X=X(l)-X(c)
now Z=squareroot(Rsqure+Xsquare)
now I=V/Z
THEN V(l)=I*X(l)
and V(c)=I*x(c)
and v(R)=I*R