Question

By answering the following questions, explain why τ=RC makes intuitive sense. Do not just say "the formula τ=RC says so" for this part; the point is to justify the formula, not to use it (nor to derive it, as properly deriving it uses calculus).

1. Explain (qualitatively) why it is reasonable for the "time to equilibrium" to increase with resistance.
2. Explain (qualitatively) why it is reasonable for the "time to equilibrium" to increase with capacitance.
3. Explain (quantitatively, using dimensional analysis) why τ=RCτ=RC is the "only option" for the timescale in an RC circuit. (You may assume that the only "inputs" are the resistance RR, the capacitance CC, and the input voltage V0V0, in this explanation.)

BACKGROUND INFORMATION

RC Circuits

In real circuits, capacitors do not instantly equilibrate - the charges take time to move. The length of time it takes for it to reach equilibrium depends on the resistance of the other components - how much they resist the flow of charges (i.e., current) through them. The larger the resistance, the slower the charges move, so the longer it takes to reach equilibrium.

This behavior can be summed up as follows.² With a resistance (with SI units of ohms, denoted by Ω) of R, the time it takes to get "close" to equilibrium³ (or time constant) is denoted by:

τ=RC(5)

An "ideal" wire should have zero resistance - it would be a perfect conductor, so charges move around instantaneously. (Real wires, combined with realistic contact resistances, can add up to ~Ω-scale resistances across the lengths of wire used in this lab.)

On the other hand, an "ideal" voltmeter should have infinite resistance, so that charges cannot "leak" through it. (A real voltmeter will tend to have ~MΩ-scale resistances.)

Answer 1

Let t be the time to equilibrium:

a)

How t varies with R while C remains constant?

Ans:

It's about either discharging or charging the capacitor (C). At a given voltage, as the resistance increases the current decreases, i.e the rate of flow of charge decreases. If the rate decreases then the time taken should increase.

As resistance increases the time taken will also increase and as resistance decreases the time taken will decrease for a given capacitor.

b)

How t varies with C while R remains constant?

Ans:

Now while charging or discharging the charge on the capacitor changes and hence the voltage changes. As voltage changes for a given capacitor the rate of flow of charge will change for a given resistance.

If the voltage decreases by , then the rate of flow of charge will also decrease and hence the time will increase. (also vice versa)

For a given change in charge, will be less for higher capacitance.

Hence, the higher the capacitance higher is the time taken.

c)

t = seconds

R = Voltage/ Amperes = (Voltage/Columb) seconds

C = Farads = (Columb/Voltage)

Hence, R*C will have seconds as units and also when t = RC, the statements in part a, part b of the answer will be plausible.