# series LC questions

P

#### PDRUNEN

Jan 1, 1970
0
Hi Group,

When the jwL is equal to -1/(jwc) at Frequency Fres for the series LC, the
circuit is a shunt and become a "short" to RF at the resonate frequency.

Given that the Inductor is on the top side and the cap is on the bottom side
with one end tied to ground potential.

If the sine source is 50Vrms with 1 ohm output and the reactance are j50 and
-j50 and I measure the voltage across the cap, would it be Zero Voltage or 50
volts are many takes greater than 50 V?

Tnx de pdrunen

A

#### Andrew Holme

Jan 1, 1970
0
PDRUNEN said:
Hi Group,

When the jwL is equal to -1/(jwc) at Frequency Fres for the series LC, the
circuit is a shunt and become a "short" to RF at the resonate frequency.

Given that the Inductor is on the top side and the cap is on the bottom side
with one end tied to ground potential.

If the sine source is 50Vrms with 1 ohm output and the reactance are j50 and
-j50 and I measure the voltage across the cap, would it be Zero Voltage or 50
volts are many takes greater than 50 V?

Tnx de pdrunen

I think you mean 50Vrms with 1 ohm output impedance

Current = Voltage / Impedance

I = 50 / (1 + j50 - j50) = 50/1 = 50 Amps RMS

Voltage = Current * Impedance

Voltage across capacitor = 50 * -j50 = -j2500

i.e. 2500V RMS 90 degrees out of phase with the current.

The voltage across the inductor is the same magnitude, but 90 degrees
out of phase the other way (+j2500)

The voltages across L and C are 180 degrees w.r.t each other - so they
cancel.