1. The problem statement, all variables and given/known data 20 bulbs are wired together in a series position. A) what happens if 10 more bulbs are added in series? B) the bulb was modified such that an additional resistor is placed across the filament (parallel to the filament). The value of the resistor R is much larger than the resistance of the filament. How does the brightness change? 2. Relevant equations V=RI P=VI 3. The attempt at a solution A) brightness decreases as the voltage is becomes smaller across each bulb hence since P=V^2/R where R is the constant resistance of each bulb, the power drops and subsequently the brightness drops. I'm more concerned about part B as I was marked wrong even though I seem right. Here is my answer: although the effective resistance of the bulb decreases, as there is exactly 20 of them hence the voltage across each of the bulb is still the same. And since each bulb contains a parallel branch of the filament and the resistor thus, the filament still receives the same voltage despite their difference in resistance. Since the resistance of the filament remains the same, that means the power developed by the bulb also remains the same. Thus the brightness of each bulb also remains the same. I don't get how this is wrong. I asked my teacher about it but she seemed uncertain about it. At first she said that the power developed as a whole is greater so the brightness increases. I explained that the resistor doesn't give out light but only heat (she agreed with that). So I explained what I wrote above and her, with a totally different answer from before, said that since the question says that the resistance of the resistor is a lot higher that means most of the current will past though the filament (short circuit in essence). But the sudden change in answer makes me feel worried. Could she be embarrassed to admit her mistake because this is my mid years paper and many students did it so it will be harder to change all of our marks or am I really wrong? Thanks for the help!