On RFID resonant circuits the Q is not usually so high that you can't place 100kohm in parallel with it. That's the first point. The second point is that the 100k could be 99k and 1k in series with the 1k connected to ground. Connect your scope across the 1kohm to do measurements.
The 1kohm and 100pF of your probe won't start to attenuate frequencies till about the 3dB point and that is: -
\$\dfrac{1}{2\pi\cdot RC} = \$ 1.6 MHz.
OK there will be attenuation of 100:1 but it may be visible. If it isn't I'd consider using a decent fast op-amp stage as a gain of ten to help out your viewing of the signal.
Start with DC batteries. Easier to understand. We'll turn them into vectors based on polarities which will help when we go to AC.
Series: \$E_T\ = E_1∡0°\ +\ E_2∡0°\$. If the batteries are identical. \$E_T\ = 2\ E_1∡0°\$. Current will be Ohm's Law, \$I\ =\ E_T\ /\ R\$. Twice the single battery current if R is constant.
Series: \$E_T\ = E_1∡0°\ +\ E_2∡180°\ =\ 0\$. Opposite polarities means no voltage or current.
Parallel: \$E_T\ = E_1∡0°\ =\ E_2∡0°\$. If the batteries are identical (same voltage, capacity, etc.) then \$I\ =\ E_T\ /\ R\$ and each battery will supply half the current to load. If the batteries are not identical in every way, current will flow from higher to lower and quickly discharge (in secondary cells).
Parallel: \$E_T\ = E_1∡0°\ =\ E_2∡180°\$. Both batteries act as loads and will quickly discharge and possibly explode or cause a fire. (Hence a lack of response to your question).
So single-phase AC.
Series: Each AC source has a magnitude and an phase angle \$V_1∡0°\$ and \$V_2∡θ\$. Now you must do vector addition on the two voltage sources to find the resultant. \$Vector\ V_R\ =\ Vector\ V_1\ +\ Vector\ V_2\$. Again Ohm's Law gives us current \$I\ =\ V_R\ /\ Z\ =\ V_R\ /\ R\$ - Assuming a resistive load. Current is also a vector. As phase angle between sources varies between 0° and 180°, current will vary from twice to 0.
Parallel: Now, assuming they are identical (same voltage, frequency, phase angle, etc.), current will flow from each source to load with each source providing half of the total.
If they are not identical, large currents will flow in the windings of the generator coils, hopefully activating protection or burning out generators.
In principle, if the main characteristics of devices are identical, you can generally parallel many electrical devices (drivers, regulators, transformers, generators, batteries) to get the same voltage and supply a proportional amount of current.
Three-phase generators can be paralleled if phase sequence, voltage levels and frequencies are the same. Generators can have different kW or kVA ratings with each supplying the same proportion of the load. 200kVA and 100kVA generators operating in parallel at 60%, would supply 120kVA and 60kVA.
A three-phase generators can be connected in wye or delta. When connecting a new generator in delta, odds are that the manufacturers instructions will state connect phase 1 to phase 2, phase 2 to phase 3. Then apply a voltmeter to unconnected terminals of 1 and 3 and measure the voltage when the generator is powered up. If 0V, phase 3 can be connected to phase 1. If one of the phases is backwards, \$2\ ×\ V_{PHASE}\$ would be applied to small impedance of series connected generators, creating large currents and quickly burning out coils.
Best Answer
That's because the model you are assuming does not well replicate an rfid receiver.
The EM wave that couples with the coil produces some varying voltage at the coil's terminals, so the voltage generator, with a series resistor if you want, should be in parallel with the coil. Now the best you can do is adding a parallel capacitor so that the LC parallel resonates at the right frequency provoking a very high voltage gain (not power gain).