This 833Ω is now in series with the 100Ω resistor, which gives a total resistance of 933Ω.Tags: Gcse English Language Coursework 2011Develop Critical Thinking SkillsProblems Solving TechniquesTemplate For Writing A Research ProposalCreative Writing Competitions For High School Students UkWalker Evans EssayBiology Coursework On BeetrootDo My Physics Homework
Notice how the 1KΩ resistor and the 5KΩ resistor are in parallel.
Therefore, we can can take the resistance of these 2 components by the formula, R2 || R3 = (1KΩ)(5KΩ)/(1KΩ 5KΩ)= 833Ω.
Adding 8.91m A and 1.78m A gives us the original 10.7m A.
Now that we know the currents we can figure out the voltage with ohm's law, with the formula, voltage= current * resistance (V=IR).
The currents subtract because they are in opposite directions.
The current going through the 5KΩ resistor is, 1.78m A 0.09m A= 1.87m A.
With this we can now calculate the total current (I= 4.6m A(98Ω/100Ω)= 4.5m A. The voltage drop across the 1KΩ resistor is, (4.6m A)(1KΩ)= 4.6V.
The voltage drop across the 5KΩ resistor is, (0.09m A)(5KΩ)= 0.45V.
At the end, once we have analyzed the circuit from each power source separately (with that one power source in the circuit and the other removed), we add up all the currents and voltages in each part of the circuit and this will equal the total current and voltage (in that part of the circuit).
So the best way to see the superposition theorem in practice is do an example. So now we'll analyze the circuit below with 2 power sources (voltage sources).