With this question, I am not too sure about the V1. Would it be the same as the 15V source on the left or is my calculation correct?
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\$\begingroup\$ V1 is determined by the 8k and 16k resistors and the voltage to the left of them. \$\endgroup\$– Andy akaSep 13, 2020 at 9:48
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1\$\begingroup\$ Um, you get a higher voltage than you put in, only with resistors. That can't be right, can it? \$\endgroup\$– Marcus MüllerSep 13, 2020 at 9:52
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\$\begingroup\$ Just like in your previous question, you're just randomly ignoring or combining resistors. You can't do that. You need to go back to the very basics of circuit analysis (you certainly learned that from some material), and only do things that are explicitly allowed. \$\endgroup\$– Marcus MüllerSep 13, 2020 at 9:54
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\$\begingroup\$ I think I know what my mistake is. The voltage in parallel should be the same. So therefore my V1 should be 15V also. \$\endgroup\$– King of ZimbabweSep 13, 2020 at 10:07
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1\$\begingroup\$ @David No, that's again ignoring a resistor. can't do that. You really need to go back to the basics; it feels like you're really just randomly hope to do the right thing, and I don't even understand how your approaches come to be! \$\endgroup\$– Marcus MüllerSep 13, 2020 at 10:33
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1 Answer
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The voltage at V1 = 8kOhm / (8kOhm + 16kOhm) * 15V = 5V .
You have the right idea with current, however, it is flowing in the opposite direction as you have shown in the 10kOhm resistor.
For the 8/16kOhm resistors, the current is : 15V / (8kOhm + 16kOhm)= 0.625A
The total current is therefore: I = 1.5mA + 0.625A
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1\$\begingroup\$ Welcome to EE.SE. For homework type questions we don't give complete solutions as this does no favour to the OP or the site (which doesn't want to become a free homework service) and note that these days the OP could be in an exam at home. Instead provide a clue or a question to get the OP thinking and as they update their question you can add to the answer. It's not how the site is supposed to work but it can be used successfully. \$\endgroup\$ Sep 13, 2020 at 10:25
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\$\begingroup\$ If you're going to hang around then note that you can use HTML entities
Ω,μ,°etc.,as well as<sup>...</sup>and<sub>...</sub>for superscript and subscript. The SI standard recommends a space between the numeric value and the units. (This is the same as '5 apples' rather than '5apples'.) \$\endgroup\$ Sep 13, 2020 at 10:28 -
\$\begingroup\$ This is not an exam, just practice. Exams would be in November at all universities in my city. To reply to the answer, I see that the voltage divider rule was used. Would the current at V1 not be the same at Ia or am I missing something? \$\endgroup\$ Sep 13, 2020 at 10:35
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\$\begingroup\$ This is my new calculation using voltage divider. imgur.com/a/Lb5Nykj \$\endgroup\$ Sep 13, 2020 at 10:46