<p>@HBOUND are you sure that the answer for the first question is that parallel bulbs are brighter? I put that answer but the thing that the sum of the currents in parallel equals the current of the bulb in series makes me thing the opposite. Can you explain the concept please.</p>
<p>@HBOUND @Andygus</p>
<p>Isn’t the brightness of a bulb determined by its power? And P = IV and the V is the same for all but I is less for the bulbs in parallel, making their power and thus their brightness less? That’s what I thought but I might be wrong.</p>
<p>^ That is correct . I agree with you</p>
<p>I thought when the switch is closed, one of the parallel bulbs is off.</p>
<p>yeah I got a for the 1st 3</p>
<p>the one in series was defn. brighter. the other one was all 3.</p>
<p>i used p=iv thing is voltage is different for series and parallel. i think voltage is higher for the parallel.</p>
<p>@HBOUND @Reda1234 </p>
<p>[erika’s</a> physics blog: DC Circuits](<a href=“http://hperikai.blogspot.com/2011/05/dc-circuits.html]erika’s”>erika's physics blog: DC Circuits)</p>
<p>I think this guy explains my thinking… Am I missing some part or is the third image on this page the exact same circuit?</p>
<p>Sorry I made a mistake in my previous calculations. if each bulb had a resistance R=1 ohm and the voltage of the generator = 1 V.</p>
<p>then the overall resistance of the circuit is 1.5 ohm.</p>
<p>I=V/R then I=2/3 </p>
<p>if we consider the parallel circuit as a single component it has resistance 0.5.</p>
<p>V=2/3*1/2=1/3 then the series light bulb has V=1-1/3=2/3.</p>
<p>if the brightness is power which equals P=V<em>I, the the power of each bulb in the parallel circuit is P=I/2</em>1/3=1/4 while the P of the series light bulb is P=2/3*I=1 watt</p>
<p>thus the Brightness of the one in series is the largest.</p>
<p>For the graph of Volts vs. Amps one where it asked you for the slope, was is 3 Ohms? (Not sure if the number is right but I definitely remember putting Ohms as the unit)</p>
<p>@NWskier</p>
<p>that’s what i put too. it was 24 volts on the y-axis and 8 amps on the x-axis.</p>
<p>what did you guys get for the one with the wire and cable, both with current going to the top of the page?</p>
<p>I think the light bulb in series should be the brightest one. Cuz its current = the sum of the other 2’s current. That was what I thought.</p>
<p>@eyeofthemonster yeah, 3 ohms. fairly easy. i am being swayed on the series circuit being brighter.</p>
<p>@dragon123, you are right. more current in series. -5 for me. i think i got 800. thank god for the curve.</p>
<p>@Eyeofthemonster . The forces of the currents are inward because theyre currents flow in the same direction. Currents with the same direction current attract.</p>
<p>You would be find. -15 is still 800. I skipped lots of EASY questions because I didn’t remember the formula. The curve is pretty generous, so -20 ( i assume lol) would be fine for me.</p>
<p>YES i got the circuit questions right haha. What about the questions about the index of the two mediums. They have a picture of refraction from one medium to another. That was tricky</p>
<p>I missed the last question about the 2 speakers. What did you guys say for the piston ?</p>
<p>@mbomb99: This is what I think. Because the light bends away from the normal, so n2 < n1. n = c/v, so as n decreases , v increases. The frequency of the light is constant regardless of the medium. So v increases indicates wavelength in n2(the shaded region) increases. I chose vA < vB , wavelengthA < wavelength B. The next question is fA = fB, vA < vB</p>
<p>@mbomb99: I chose piston ( pull in) , what was the other one? I don’t remember.But I chose E</p>