Conceptual Circuit Problem

Imagine the situation where the following is connected with the capacitor uncharged at the instant of connection. Imagine the situations of the instant of connection versus the behavior of the circuit after one waits for a very long time and the capacitor is fully charged and will accept no more charge.

R1 C = 200 mF

R2

R3

ε

A 200 mF capacitor is one such that if you know there is a 1 V span in the gap from plate-to-plate, then that means those plates contain +200 mC and -200 mC. If 5 V spanned the gap between the plates, then the charges on the plates would be 5 times as much. The voltage in the gap is a result of the charge separation. The voltage does not cause the charge separation of + and – 200 mC.. The voltage exists because there is electric field between the capacitor plates, and it should be obvious by now that E field is the result of the oppositely charged plates. The voltage does not cause the charge. The definition of capacitance is being described in this paragraph. The illustrations make it fundamental that Q = C(ΔV), and C is the number of Farads being referred to. 200 mF is an example of a C value, a capacitance. 200 mC is not an example of a C value, because mC is not the unit of capacitance. mC is a unit of charge, so 200 mC is a Q value. Anyone reading this paragraph is supposed to have figured out in the margin by now that the definition of Farad is that Farad is a unit composed of Coulomb/Volt. C is not a unit; it’s a category of measurement, and it’s called capacitance. C is not measured in C. C is a unit; it is the metric unit of the category called charge, a category usually symbolized with a Q. Q is measured in C, and that is called a Coulomb. C is measured in F, and this category name (NOT A UNIT NAME) C is called capacitance. C is measured in C/V, and therefore capacitance is not measured in Coulombs, and therefore capacitance can’t mean the same thing as charge. Capacitance is measured in F, and F is identical to C/V. If this has potential to be confusing, then you don’t play that game and never call capacitance’s units Farads and simply plug things into Q = (200 mC/V)(ΔV), using the full language strongly. Either way, it the units don’t cancel, you don’t know what you’re doing.

Problem: Let the emf of the power supply be 1.5 V. Choose all three resistances such that the current coming from the power supply immediately upon connection is 400 μA and then after waiting a very long time, that same current approaches the asymptote of 220 μA.

Key:

Even though infinite answers are possible, the following answer must be understood.

Answer: R2 and R3 must add up to be 6818.2 Ω.

Based on this value of 6818.2 Ω, do more problems:

2. If I want R3 to be 2000 Ω, what does R1 have to be?

3. What does the graph of the current out of the battery versus time look like? (It’s concave up with an ever-less-steep slope as time approaches infinity. It’s starting value on the I intercept is 400 μA, and the value it approaches flat and asymptotically is 220 μA. Draw this graph.

4. On the graph you drew, mark a time T when the current coming out of the battery is 250 μA. At this instant, solve for the charge on the capacitor.

Key:

2. 2748.2 Ω

4. 0.177 C

When doing #4, your strategy was to use fundamental Kirchoff’s Law Rules with no shortcuts, and through these methods, you get that the capacitor voltage is 0.883 V at the instant T in question. You then apply Q = C(ΔV) to state the charge.

0.883 V was the capacitor voltage at the time T. What is the capacitor voltage at the infinite-time steady state anyway? Derive it.