Timeline for Mesh Analysis for circuit having Current Source
Current License: CC BY-SA 3.0
12 events
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| Apr 8, 2016 at 4:59 | comment | added | elecman | okay..thanks a lot..can you please ping me when you are online... i dont want to post my mail address for a long time here...I will post the mail address here when you are online.. | |
| Apr 6, 2016 at 15:21 | comment | added | dsmith | @elecman, I got what you want, but can I send it to your email? | |
| Apr 5, 2016 at 10:17 | vote | accept | elecman | ||
| Apr 5, 2016 at 9:54 | comment | added | elecman |
okay..thanks.Please see aries.ucsd.edu/najmabadi/CLASS/MAE140/NOTES/analysis-2.pdf Page 34 Here they are calculating i by saying that The problem unknowns, i and v can now be found from the mesh currents: i = i 1 − i 2 =1A .Which mesh equation are they referring? cant it be just the opposite i2-i1 from the mesh 2?
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| Apr 5, 2016 at 9:43 | comment | added | dsmith | @elecman, Yes. The 12 ohms resistor is split between two mesh loops, thus the current going through the 12 ohms resistor is affected by the two mesh currents. If the 12 ohms resistor was on the wire where the 2 amps current was, then it would only be affected by one mesh current which will be I1, which would have been the only current going through it. | |
| Apr 5, 2016 at 9:28 | comment | added | elecman |
okay... just because i1 circulates through mesh1 we cannot say that the current entering the 12ohm resistor is i1 right?
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| Apr 5, 2016 at 9:11 | comment | added | dsmith | @elecman, you have to do a KCL at the 12 ohms resistor's top node. Lets assume that currents entering that node are -ve and currents leaving that node are +ve, you have: -I1 + I2 + I12 = 0. Assuming I12 is the current going from top to bottom in the 12 ohms resistor. | |
| Apr 5, 2016 at 8:58 | comment | added | elecman | okay.If we are asked to find the current entering the 12 ohm resistor,is this same as i1? | |
| Apr 5, 2016 at 7:31 | comment | added | dsmith | @elecman, I1 is the mesh current in Loop 1 which means that it goes through ALL WIRES that make up Loop 1. Looking at Loop 1, there is only one current source going through three wires (which is 2Amps). Since I1 is the mesh current going in the same direction as the 2Amps and there is only one current source going through those wires, I1 is 2Amps. | |
| Apr 5, 2016 at 4:40 | comment | added | elecman | How do you say i1=2A without any calculation? There are other circuit elements in the same mesh.. can we assume like this?Plus 2A was not given in the question.Can this question be solved without it? | |
| Apr 5, 2016 at 4:29 | comment | added | elecman | Thanks again for the detailed answer.Sorry for the late reply.. i just came back online now..I will take a look at your answer.Thanks again :) | |
| Apr 4, 2016 at 12:45 | history | answered | dsmith | CC BY-SA 3.0 |