View Full Version : please help: #34 in PS on Full Length 3 in Kaplan
aviary17 07-20-2008, 05:40 PM could you please check this problem and explain it to me? it is my understanding that the voltage drop across every resistor is a parallel circuit is the same V, but Kaplan's explanation and answer (C) seem to imply otherwise. Thanks for your help.
futuredoctor10 07-25-2008, 03:12 PM Hey aviary17... I am taking FL3 soon so once I do I will let u know about this prob :)
minhaj 07-25-2008, 03:43 PM The question says there are two resistors with a resistance of 10 ohms and the voltage is V.
Current (I) going through R1 = V/10 ; Current (I`) going through R2 = V/10.
Now when R1 is taken off of the circuit, we only have R2 = 10 ohms. The voltage of the battery is the same. Hence, the Current going through R2 = V/10, which is same as the current going through R2 when R1 was a part of the circuit.
The current going through the circuit, however, is reduced by 1/2 because now the total resistance is doubled.
Hope that makes sense
aviary17 07-27-2008, 03:35 PM Min, thank you for your response, but I am wondering if you are talking about the same problem... with the speakers in the circuit?? Let me know - Thanks!
And futuredoc, I'll be waiting for what you have to say, thanks for your help.
murdoc9 07-27-2008, 04:10 PM the voltage drop across each element in a parallel circuit is the same. in this problem, "V" is the potential difference across the battery, but not necessarily the potential difference across the parallel circuit. (we're not told that the voltage drop across the parallel circuit is exactly that of the battery.) So a more accurate way to determine the voltage drop is to look at the current through a known resistance (the current through R). We can know for certain that this voltage drop will be the same for the speaker since they're in parallel.
aviary17 07-27-2008, 04:39 PM Murdoc, why would you ever assume that the voltage drop across the resistor is not the same as that of the battery? I thought in parallel circuits, the V of the battery is the same everywhere...
murdoc9 07-27-2008, 05:00 PM i guess they expected us to assume there was extra resistance in the circuit- ie from the ionizing cylinder, wire resistance, etc. i agree it's kinda shady, but the surest way to know the V of the parallel circuit is to look at the current flowing through an element of that circuit with a known resistance.
futuredoctor10 07-28-2008, 05:10 PM Yeah I think they must have wanted us to assume the ionizing cylinder will affect the circuit, so in order to determine the Voltage across the speaker, we had to know the voltage across the resistor (by knowing "R" and current I)
Although I got it right I think this is a very questionable/arguable question. The battery appears to be in parallel with the resistor and the speaker, which makes one think the voltage difference across all three would be equivalent.
futuredoctor10 07-30-2008, 07:00 PM So I've asked someone else about this question. We think it is the diagram that helps set up the question - the ionizing metal cylinder and wire present is affecting the circuit and voltage. Therefore, the battery is NOT in parallel with the resistor R and speaker (or we cannot assume it to be parallel).
This makes "D" [no other info required] incorrect, and "B" right- by knowing I we can calculate voltage across the resistor which must be equal to the voltage across the speaker (since resistor and speaker are in parallel)
aviary17 08-02-2008, 04:05 PM Thanks!
StayingSteady 08-14-2008, 07:54 PM i guess they expected us to assume there was extra resistance in the circuit- ie from the ionizing cylinder, wire resistance, etc. i agree it's kinda shady, but the surest way to know the V of the parallel circuit is to look at the current flowing through an element of that circuit with a known resistance.
omg this is question is so dumb. what a dumb question. so dumb.
IkerUnzalu 08-19-2008, 08:46 PM Keep it simple. Resistors in parallel decrease the resistance and resistors in a series increase resistance. so when you take one resistor in parallel out then the circuit is in a series which according the Mr Ohm will decrease the current that goes through it.
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