a simple RLC series resonant circuit issue

[sj95]

*series resonant
.options fast

Vin nco gnd SIN (0 1 300 0u)

R nco a 20
C a b 5.897617363u
L b gnd 0.04777070064

.TRAN 1u 1000000u

.print tran v(vin,a) v(a,b) v(b) i(vin) i(r) v(nco,a) v(a,0)
.END

Hi all:

I make a simple RLC series resonant circuit, it shows jXL= -jXC at 300Hz , so the imaginary part of Z =0 , the V(a) no voltage drop, V(a)=0v.
The peak Vin voltage is 1v ,then the i(Vin) is 50mA

If I change the vin to
Vin nco gnd SIN (1 1 300 0u)
The peak Vin voltage is 2v

But , the V(a) is 1v at 300Hz now. And the i(Vin) is still 50mA

Why ? if jXL= -jXC , the imaginary part of Z is 0 , should the v(a) voltage be 0 ,too ?

Any correction and suggestion is welcome.
Thank you very much.

 

[Debdut]

You can think of it like this.
Break the vin in 2 parts - sine and dc
Capacitor blocks dc. If the dc is 1V, then voltage at 'a' is to be 1V as capacitor is blocking any dc current, so no drop across resistor.