How is current affected by a charging capacitor in parallel to a resistor?

Question

Imagine a circuit with a battery followed by a lightbulb and an uncharged capacitor in parallel. Next to it is an open switch. When I close the switch completing the circuit, does the brightness of the lightbulb change over time due to the capacitor charging?

My reasoning is that the brightness must change slightly as an uncharged capacitor initially does have current flowing through its path. In an uncharged state, a capacitor can be equated to a resistor (air over a short distance still having a high resistance, of course).

So for the while the capacitor is not yet fully charged, it is lowering the effective resistance of the entire circuit thus allowing more total current to be generated with our constant battery EMF. This extra current will allow more power usage for our lightbulb by the equation $P=I^2 R$ .

I had a question like such in a HS class and the teacher's explanation was that the brightness should not change because the potential difference across the lightbulb is always the same. I do see even with my reasoning that perhaps the extra current generated would have to be split across both paths meaning maybe that the lightbulb always has the same current flowing through it, regardless. However, for my own understanding I would like to see a more detailed response as my teacher told me as well that the current should be constant. Maybe it could be approximated to be unaffected by our capacitor charging, but I am unsure if he is totally right.

Answer 1

The total current of the circuit does indeed change but as the lightbulb always remains across a constant voltage and has a constant resistance it will always pull the same amount of current from the circuit.

Kirchhoff’s Junction Rule Applies: $\frac{V_T}{R_T}=\frac{V_L}{R_L} + \frac{V_C}{R_C}$. The constant value $\frac{V_L}{R_L}$ indicates current drawn through the lightbulb. This does not change even if we vary the total resistance by adding parallel components with resistance. Of course adding straight wire in parallel would just make the voltage drop value to be zero and short the circuit.

This can also be intuitively explained as you have a material with some resistance which is able to supply a current through its path because of a potential drop and electric field through it. Adding components in parallel does not alter those characteristics. It still is the same material with the same voltage. Why should it get different current through itself?

Answer 2

Since you have defined your circuit using physical devices, and you seem to be asking a question about the same, the difficulty you are having is caused by your neglect of the resistances of the battery and the connecting wires and switch. Since these resistances are all in series with the light bulb and capacitor, the expected variable brightness of the light bulb will occur if physically tested, and the reduction in current through the bulb at turn-on can be calculated if the actual resistances be known.

Answer 3

The simplified model of a battery as a constant voltage source and a capacitor as a perfect capacitance falls apart when you close a switch connecting one to the other.

The capacitor draws current proportional to the rate of change of voltage, so for a perfect capacitor to change voltage instantaneously the current would have to be infinite. A real battery cannot provide infinite current. Even if you short it (not a good thing to do) the current will be limited. An AA cell (or 8 of them in series to give us 12V) might provide a couple of amperes into a short circuit.

If we improve our battery model slightly by including a fixed internal resistance (a crude model, but good enough for a first cut) the resistance in the case of 8 AA cells in series would be something like 6Ω.

Let's further model the light bulb as a 100Ω resistor (again, a bad model, since incandescent bulbs are much lower resistance when cold, but good enough for starters).

When we close the switch some of the current will go through the light bulb and some will go through the capacitor. The latter current will decrease over time.

In the case of a constant resistance bulb, one can show that the capacitor current will decrease exponentially with time, with a time constant $\tau = RC $ where $R =\frac {Rbat\cdot Rbulb}{Rbat+Rbulb}$. After a long time, the bulb current will be 12V/106Ω or about 113mA.

Below is a simulation of the situation. The trace shows the bulb current vs. time.

The time constant is calculated to be about 0.566s and the simulation confirms that. (After one time constant we get to about 63% of the final current).