# For a real battery, why is two identical bulbs in parrallel slightly more dim than just one in series

Image for reference:

\begin{align} &r = 0.8\,\Omega\\ &\mathscr{E} = 1.5\,\mathrm V\\ &A = B = 6\,\Omega. \end{align} The current when the switch is open, the only other resistor is the A bulb, so total resistance is 6.8 ohms, but when it's closed the total resistance should be 3.8 ohms (2 6 Ohm resistors in parallel = 3 ohm $$R_{\mathrm{eq}}$$), but that seems to not be the case.

Why would the current be less when it is closed? The way the question is asked it seems like it should be less, but the math makes it seem like more.

Here's a hint: if $$r = 0$$, closing the switch does not change the current through bulb A and doubles the current through the battery.
Try finding the voltage across bulb A when the switch is open and then when the switch is closed as a function of the internal resistance $$r$$. You should be able to do this using voltage division and the information you've already given.