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Electrics

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Thread replies: 12
Thread images: 3

File: 1489400735311-40610893.jpg (1MB, 3120x4160px) Image search: [Google]
1489400735311-40610893.jpg
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Been given this question, is there a mistake in the diagram or am I being a stupid cunt? Are all the cells the right way round?
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>>8744167
The voltage sources as drawn add to zero volt. If the 0.2A thing is a current source you know the loop current. B, C and D are in parallel and each has 60Ω because 60||60||60=60/3=20Ω, the combined resistance. Since current follows conductance each carries one third of the loop current. Resistance of A comes from P=I^2*R. Loop resistance is given as 65Ω=20+20+r/4 and the pd is 0.2A*r/4.
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>>8744167
It's just a regular old Kirchoff's laws task - i.e., completely abstract in order to teach you how to solve general circuits. Should have paid attention in class, dumbass.
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>>8744409
Thanks basedanon. Yeah I should have paid attention
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File: pd.jpg (3KB, 400x29px) Image search: [Google]
pd.jpg
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>>8744167
What does pd stand for? Could be potential drop (5V) or power dissipation (1W).
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>>8745516
It stands for potential difference
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>>8747199
So either 12V or 2V depending on the unknown direction of the current?
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>>8747482
Thanks. Do you mind explaining how you got that?
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>>8747766
You can reduce the four parallel cells and their internal r to one 7V source and one series resistance of r/4. The voltage is compensated by the single 7V source of opposite polarity and the current is controlled by the 0.2A source alone. B,C,D in parallel have 20Ω, A has 20Ω (from R=P/I^2) and the total resistance is 65Ω, which leaves 25Ω for r/4 (or r=100Ω). If you switch off the current and measure the voltage across the four-cell assembly you would get 7V. With the current of 0.2A you have 0.2A*25Ω=5V across the r/4 resistance. The direction of the current is not specified so you either get 7+5=12V or 7-5=2V across the parallel combination of cells.

Another scenario would be that the single 7V cell has been drawn with the wrong polarity and 0.2A is the measured current. In this case you have a 14V source, the current flows ccw and the potential difference across the parallel combination is 2V. All other values remain unchanged. Or so I think.
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>>8744167
Yes, you're going to see a lot of that when you get to modeling diodes.
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>>8744167
wye/delta
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File: c_loop.png (21KB, 291x462px) Image search: [Google]
c_loop.png
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>>8748010
Simplified original (current set cw) and alternative version (current ccw).
Thread posts: 12
Thread images: 3


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