<p><a href="http://www.act.org/aap/pdf/preparing.pdf%5B/url%5D">http://www.act.org/aap/pdf/preparing.pdf</a></p>
<p>The problem is number 17 on page 47. I have no prior physics knowledge, and I don't see how to solve this problem without it. Anyone care to help?</p>
<p>luckily, i know how to do this without looking thanks to physics... but without that, i'm sure this could be hard... the answer is in the passage though... look on page 46, figure 1 (and the short paragraph right before it that is right under the words 'Passage III")... it tells you that the voltmeter (circle with a V in the middle) measures voltage... the || is a capacitor and the zig-zag line /\/\/\ is a resistor... if the (V) measures voltage and the /\/\/\ is the resistor, then it makes sense that in order to measure the voltage across the resistor, you put the voltmeter at a point before and after the resistor to only include the resistor when calculating the voltage... so A is the answer</p>
<p>wow... i actually messed up on that problem on the actual ACT when i took it. it was the same one as ur practicing LOL. Now after being vetted with ACTs, i realized it's easier than i thought. Yea A is the answer.</p>
<p>But couldn't the current go to the voltmeter before going to the resistor in figure A? (Starting from the battery, going down, right, then up)</p>
<p>for one, the current will never come from the negative part of the battery (the shorter line)... it always flows from the positive side (the longer side)</p>
<p>but that's not the point.... ok, here's a physics lesson (for future science sections)... the voltmeter does provide an alternate path for the current, yes... but it is set up in parallel to the resistor.... in parallel, the voltage stays the same no matter what path it takes and the current splits up (so it's weaker on those two paths than anywhere else in the circuit)..... but since the voltage stays the same no matter which path you take, the voltmeter will read the same voltage that will go across the resistor since the voltage is identical across both pathways.</p>
<p>it's a screwy concept, but that's it... electricity was my forte last year in ap physics b</p>
<p>It's A. We want to see how many volts were "absorbed" by the resistor, the voltage drop across it. To do that, we just place a voltage reader before the resistor and after it.</p>
<p>^^ pretty much, there's a shorter explanation.. i was just explaining how it worked</p>