- #1
Basic Electricity: Learn Basics of Electricity
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Homework Help Overview
The discussion revolves around basic electricity concepts, specifically focusing on calculating total resistance in circuits with resistors in series and parallel, as well as understanding current flow in a circuit. Participants are exploring the implications of using various formulas and the behavior of electrical components under different conditions.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- Participants are checking calculations related to resistors in series and parallel, questioning the correctness of their answers and the application of formulas. There are discussions about the implications of shorting a battery and the relationship between current and total resistance in a circuit.
Discussion Status
Some participants have provided hints and corrections regarding calculations, while others are still seeking clarification on their understanding of current and resistance. Multiple interpretations of the problems are being explored, and there is no explicit consensus on all points raised.
Contextual Notes
Participants mention potential confusion regarding the version of the problem presented and the assumptions made about battery ratings and circuit behavior. There are indications of missing information that could affect the calculations and understanding of the problems discussed.
- #2
Doc Al
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The first problem (resistors in series) is done correctly, but the second (resistors in parallel) is not. Recalculate the total resistance in that second circuit. Hint: When resistors are in parallel, the total resistance will be less than that of any single resistor.
- #3
andrevdh
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The total resistance of resistors in parallel can be calculated with
[tex]\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]
this can be changed to
[tex]R_p=\frac{R_1R_2}{R_1 + R_2}[/tex]
[tex]\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]
this can be changed to
[tex]R_p=\frac{R_1R_2}{R_1 + R_2}[/tex]
- #4
Energize
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andrevdh said:
When I do that I get 6x2 = 12ohm 12ohm/8 ohms = 1.5 ohms, is that right?
- #5
andrevdh
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Yes. That seems to be it.
- #6
Energize
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Am I right in thinking that if you have a 12v battery and short it the current would be very high?
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- #7
andrevdh
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Not neccessarily. The amp-hours rating is a complicated affair. It is the amount of current it can supply (chosen for "normal" [designed] operating conditions) for a certain set period (I think it is normally for 20 hours, but I might be wrong) before the battery is completely drained and dies out. So you might get more out of it for a shorter period but only to a certain limit (everyone has his limitations!).
- #8
- #9
Energize
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Anyone?...
- #10
Doc Al
Mentor
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Rethink your answer for C (the total current).
- #11
andrevdh
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The current through the 2 ohm resistor is the current through all of the circuit since
[tex]I = I_1 + I_2[/tex]
So [tex]I[/tex] will be be determined by the total resistance of the circuit and the voltage of the power supply.
[tex]I = I_1 + I_2[/tex]
So [tex]I[/tex] will be be determined by the total resistance of the circuit and the voltage of the power supply.
- #12
Energize
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Ok we went through that question today and I got A-F right but G was 0.9A.
- #13
CPL.Luke
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12/5 does not equal 1.2, maybe you sent us the wrong version of the problem, or the teacher did it wrong
- #14
Energize
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Ah sorry I put 12v instead of 6, doh.