Few Questions on source tranformation

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When different voltage sources are in series, the resultant voltage is the sum of the sources, considering their polarities, rather than simply picking the smaller one. The initial conversion of 1A and 2 ohms to 2V is incorrect as the resistance does not disappear; the solution likely mislabels a resistor. The confusion regarding the disappearance of resistance arises from the parallel configuration of resistors, which can be replaced by an equivalent resistance. The discussion highlights the importance of proper labeling and understanding circuit configurations. The thread suggests a need for relocation to the homework sub-forum for better assistance.
clurt
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It's question 1b) Please view attached

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1. Just to clarify, when different voltage sources are in series you pick the smaller one?

2. First step, they turned the 1A and 2ohm into 2V, why does the resistance disappear? Then between the last two steps the 12/7 ohm resistance remains with 18/7 voltage. Why does one resistance disappear while the other doesn't?

Thanks on advance
 
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clurt said:
It's question 1b) Please view attached
It would have been better if you'd cropped the attachment to just the relevant block in post #1 and in the process made it more legible.

1. Just to clarify, when different voltage sources are in series you pick the smaller one?
No! The 3v comes about because two sources are in series, +6v and -3v, so the resultant voltage is their sum with polarities taken into account.

2. First step, they turned the 1A and 2ohm into 2V, why does the resistance disappear?
It doesn't. The solution is in error. I expect the right side 2 ohm was probably intended to be labelled 4 ohms.

Then between the last two steps the 12/7 ohm resistance remains with 18/7 voltage. Why does one resistance disappear while the other doesn't?
Nothing disappears. The two resistors are in parallel, so they can be replaced by an equivalent one of the value equal to that when two resistors are in parallel.

Hope that helps.
 
Last edited:
This thread belongs in the homework sub-forum. You almost certainly would have received prompt assistance had you listed it there.

Could a mentor kindly move this to the appropriate forum ...
 
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