Kevineamon Posted March 29, 2017 Share Posted March 29, 2017 Hi guys I'm new to the forum. I hope this is in the correct area of the forum. Struggling with a few problems on capacitance. Here's the question. Determine CT for the circuit below. So I'm thinking this is the equivalent capacitance. Now those three capacitors in uF. I'm thinking they should all be in parallel in which case they get added. I think... Then the CT for them should be 0.111uF 10000pF I've worked out as 0.01uF So 0.111 in series with 0.01... looks like to me...? [math]1/((1/0.01)+(1/0.111)) = 9.17*10^{-3}[/math] Problem is I have 0.121uF as the answer from the lecturer. Link to comment Share on other sites More sharing options...
Endy0816 Posted March 29, 2017 Share Posted March 29, 2017 (edited) Hi there! Welcome to the forums. The 10,000 pF capacitor is also in parallel. Edited March 29, 2017 by Endy0816 1 Link to comment Share on other sites More sharing options...
studiot Posted March 29, 2017 Share Posted March 29, 2017 (edited) Good evening Kevin, and welcome to SF. You certainly seem to have got into a tangle with this one,despite a lot of correct thinking. Is it homework? We have special rules about homework help. http://www.scienceforums.net/topic/75772-read-this-before-posting-in-homework-help/ Anyway Yes the first step is to convert all to the same units and 104 picofarads equals 104 * 10-12 = 10-8 = .01*10-6 farads or 0.01 microfarads. I laboured this because you seemed uncertain. All four capacitors are actually in parallel. If you look carefully, you can see that each capacitor has one connection directly to each terminal. A good way to see this is to delete three of the capacitors, but leave all the wiring intact. It is then easy to see that the remaining capacitor is directly connected across the terminals. So you need to simply add up their values as you correctly surmised. So, working in microfarads, we have 0.1 + 0.01 + 0.001 + 0.01 = 0.121 microfarads as your lecturer said. As another tip for a more complicated circuit. The right hand two capacitors form an easily recognisable parallel pair So combine these first to one equivalent capacitor before dealing with the crossed over ones. Then you can see that this equivalent capacitance is connected across (ie in parallel with) whatever the net result of the crossed over capacitors is. In this case they are also in parallel but they may not be for a more complex circuit. So you work your way through in simple steps, reducing the complexity wherever you can. It takes some practice to sort out when something is in series and when it is in parallel. Two further tips. Label your diagrams clearly and unambiguously so C1, C2, C3 and C4. This makes it so much easier to refer to the elements when someone is discussing them. Secondly this site allows you to write exponents directly using the superscipt and subscript icons in the entry box toolbar. That is such a boon. Edited March 29, 2017 by studiot 1 Link to comment Share on other sites More sharing options...
Kevineamon Posted March 29, 2017 Author Share Posted March 29, 2017 Thanks guys ! I see the parallel now. Silly me. No it's not homework as such. Well I guess it is... I'm just studying I have tests in a few weeks. Also should I have posted this in the homework section? Link to comment Share on other sites More sharing options...
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