Which bulb is brighter :) b]1. The problem statement, all variables and given/known data[/b] There is a circuit with a voltage V, and 3 identical bulbs of resistance R. B and C are parallel to each other, and A is in series to them. what happens when bulb c is removed? student one: I think that bulb B will get brighter. Bulb B used to share the current with bulb C, but now it gets all the current. So bulb B will get brighter. student 2: I don't think so. Now there aren't as many paths for the current, so the resistance in the circuit has increased. Since the resistance in the circuit has gone up, the current in the circuit decreases. Bulb B will get dimmer. Which student is correct? 2. Relevant equations 1. Sum of all current into a node is the sum of the current out of the node. 2. The sum of voltage on a closed loop is equal to zero. 3. V=IR 3. The attempt at a solution Well I agree with student one in the part where bulb B doesn't have to share the current with bulb C anymore so the amount of current going to B is increased, but student 2 makes a great point too. He says that because bulb C is gone that the amount of resistance in the circuit is increased. So from here I made the resistors into equivalent resistors in order to find the current that would be going through B in both cases. The first case I get the equivalent resistor to have an Resistance=(2/R)^(-1)+R=3/R So using V=IR, I get the equivalent I to be equal to 2V/(3R). so that would mean that to current going to B individually would be half... V/(3R). The second case without C I Get an equivalent resistance to be R+R=2R, and so the I=V/(2R). Therefore when C is removed, the current also goes up for B making B brighter, and makeing student 2 wrong. Is this correct reasoning? I am a still a little uncomfortable with circuits.