In the arrangement shown in the figure below, a potential difference [tex]\Delta[/tex]V is applied, and C1 is adjusted so that the voltmeter between points b and d reads zero. This "balance" occurs when C1 = 3.00 uF, C3 = 8.00 uF and C4 = 15.0 uF. Calculate the value of C2. I think the key here may be understanding the relationship between C1 and the voltmeter. Vbd is the cusp of this problem, I'm sure, but I don't quite get the relevance of it yet. I've looked at this a couple of different ways so far. First that C1 and C4 are parallel, and their respective capacitances can be added together to get C', their equivalent capacitance. I assumed a similar thing can be done with C2 and C3. Since the voltmeter reads 0... well first of all, bd is parallel with the plane of the power supply - you would expect a voltage ACROSS the power supply, but not along it, so I'm not sure what this tells me really. Since there is no voltage here, of course there will be no charge associated with it either. I'm encouraged to speculate the potential drops across C1 and C4 are equal to the sum of the potential drops across C2 and C3 - but I would have assumed this anyway without the voltmeter being there, reading 0, because their respective Ceq's are in series with one another. Oof. I really have nothing on this problem - please help!