what will be the resistor to find the time constant?
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 &ndash; 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$$ \tau = R \times C= \frac{L}{R}$$