Hello! I'm having a bit of a difficulty trying to solve this exercise mostly because of the resistors. So,when short circuiting the voltage source we have the 20ohm in parallel with 40 ohm right?But what about the other resistors are they in parallel or in series with the result 20//40?The exercise tells to determine the voltage with superposition in the 40ohm resistor. And what about if we open circuit the current source?Are the 2.5 ohms and 10ohm in series with each other? What kind of process should I adopt to solve for the voltage in 40ohm resistor?Currents in equal currents out? Any help is really appreciated!
Have you read about the superposition theory at this link? http://www.allaboutcircuits.com/vol_1/chpt_10/7.html It helps when you redraw the circuits at each step of your analysis. As far as your implications of currents in equals currents out ... KCL ... remember, KVL and KCL are laws, not theories. All laws apply.
20 and 40 wud be in parallel in case voltage source is killed, but other resistance wud not be in parallel or series ,some wud form a delta or star configuration,however 2.5 ll ( 20ll40 + 40 in series). when killing voltage source solve by using kcl and express current in terms of voltage at node and reference (see attachment) you will come up with equations like (v(a)-v(0))/2.5-(v(a)-v(b))/10 = 6A form other equations using kcl when current source is removed: 2.5 and 10 are in series combination is again ll 40. this combn is then in series with 20 ohms. calculate total current provided by voltage supply by v=ir then u can use a lot of ways to find current in 40 ohms resistor or voltage across it. like find voltage drop across 20 and then remaining voltage is available across 40. add algebraically the voltages obtained across 40 ohms resistor by above two steps. there are lot other ways to get to the answer try solving it with other ways.
I have one question,why is the minus before parenthesis in - (va-vb) Would the other equation for KCl be: -((va-vb))/10=-((vb))/40 - ((vb))/20
sorry the currents are in leaving direction i was being a little casual in writing eqn as it was an example eqn. the two currents shud sum up to 6 a in that case. abt the other eqn u r correct again. current thru 10 shud equal current leaving thru two resistances.