pedro_carlos Posted April 3, 2013 Share Posted April 3, 2013 What is the current that runs through the circuit below: Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted April 3, 2013 Share Posted April 3, 2013 If this is homework, we're not just going to do it for you. Have you tried to solve it? Where did you get stuck? Link to comment Share on other sites More sharing options...
pwagen Posted April 3, 2013 Share Posted April 3, 2013 Use one of Kirchoff's laws to solve it. I can't look it up properly now, but you might or might not have to apply it to all points where the currents would "meet". http://www.tpub.com/doeelecscience/electricalscience261.htm Link to comment Share on other sites More sharing options...
elfmotat Posted April 3, 2013 Share Posted April 3, 2013 This is a basic Ohm's Law problem. Show us what you've done so far and we'll give you some hints if you need them. Link to comment Share on other sites More sharing options...
pedro_carlos Posted April 3, 2013 Author Share Posted April 3, 2013 10-5-6-7 = 6 I + 8 I + 1 I +3 I + 4I + 2I -8 = 24 I -8/24 = -0,333333333 I Link to comment Share on other sites More sharing options...
imatfaal Posted April 3, 2013 Share Posted April 3, 2013 (edited) 1 pick a direction and pick a start finish point 2 If you go through a battery -ve to +ve (small side to big side) add the EMF if you go through +ve to -ve subtract the EMF 3 If you go through a resistor add IR 4. that equation should equal zero (you are back to the same point from where you started from so there can be no overall difference in voltage) - solve for I. If I is negative it goes in the opposite direction to that whcih you chose, if positive it goes in the same direction Seem a bit long-winded but it avoids putting the wrong sign on a EMF (hint). And it works for circuits with more than one loop - with a little bit of extra effort Edited April 8, 2013 by imatfaal sign mistake Link to comment Share on other sites More sharing options...
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