but I think the current caused by this voltage will be proportional
to the resistance values and so will the power dissipated by this
Let's work it out and see. We assume that the source is disconnected at t=0.
For RL and RC circuits with initial energy, i.e., there is an initial current through the inductor or an initial voltage across the capacitor, the current is given by:
\$i(t) = i_0 \ e^{-t/\tau} \$
For the RL circuit:
\$\tau = L/R, \ i_0 = i_L(0) \$
For the RC circuit:
\$\tau = RC, \ i_0 = v_C(0) / R \$
Now, let's calculate the power associated with the resistor R:
\$p_R(t) = i^2(t) R = i^2_0\ e^{-2t/\tau} \$
For the RL circuit:
\$p_R(t) = i^2_L(0)\ e^{-2t/\tau}\ R\$
As we expect for the RL circuit, the power is proportional to the resistance R.
For the RC circuit:
\$p_R(t) = \dfrac{v^2_C(0)}{R^2}\ e^{-2t/\tau}\ R = \dfrac{v^2_C(0)}{R}\ e^{-2t/\tau}\$
So, for the RC circuit, the power is inversely proportional to the resistance R.
How to intuit this without working through the math? Note that the larger the power, the sooner the initial energy is dissipated, i.e., the smaller the time constant.
Now, note the formula for the time constant. For the RL circuit, a larger R gives a smaller time constant while, for the RC circuit, a smaller R gives a smaller time constant.
In a perfect world, the capacitor would charge instantly. This is clear from your equation: the charge time is
$$
t \approx 5RC
$$
so if \$R = 0\$, then \$t = 0\$.
However, batteries are not perfect voltage sources. They have an effective resistance, which is on the order of 1 ohm, so the time to charge your capacitor without a resistor is approximately
$$
t_{real} \approx 5C
$$
This resistance depends on what type of battery, how dead the battery is, etc... so this is only a rough estimate.
Best Answer
To simply understand the circuit you can locate a proper ground point to use as a reference for calculations and transformations of the circuit. In case of the oscilloscope, you can take one of it's connectors as the ground and split the circuit to make it easier to digest.
Given the oscillograph connections, you can transform the circuit to look like this. Does it make more sense this way?
simulate this circuit – Schematic created using CircuitLab
Try simulating it and measure the time constant from the graph, then calculate what it should be. The time constant tau is the time which it takes for the voltage to rise or fall to $$V_{initial}+(1-\frac{1}{e})\times(V_{final}-V_{initial})$$ where \$V_{initial}\$ is the stable voltage before the switching and \$V_{final}\$ is the voltage after the switching. So if before the switch it was 0 and after the switch it is 1, the time constant will be the time it took for the voltage to rise to ~0.63V.
Try replacing the capacitor with an inductive coil and see what happens in the time simulations to answer your last question thoroughly. Remember that$$ \tau = R \times C= \frac{L}{R}$$