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

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.

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.