At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
There are several ways to solve this problem, but I'm assuming you're doing it without using mesh current or node voltage. You can calculate the voltage between R1 and the 2Ω resistor by simplifying those 4 resistors. The 10Ω || 5Ω = 3.33Ω. 11.33Ω || 3Ω = 2.373Ω. Then calculate the V across R1 with \[ Input V\times \left(Parallel R \div Total R \right)\]\[10V\times\left(2.373\div4.373\right)\] From there it is easy to calculate the current through R1 by ohms law.
i think you could first calculate the total resistance: Req = 2+ ((10||5)+R2)||R1) then calculate Iin I = V/Req then calculate the voltage drop on the 2ohm resistance, so the voltage across R1 is V - Iin * 2 = Vr1 and finally by ohm's law: Ir1 = Vr1 / R1
I like using a spreadsheet. It helps with error checking also. To stimulate some peoples greater interest: When analyzing AC cirduits it generated results that I thought helped me to better tune an equalizer to generate a flatter curve response by assuming that it was a set of parallel filters. The graphing of results of the frequecy amplitudes was of the greatest interest.
The 5 ohm and 10 ohm resistors are in parallel and can thus be replaced by a single resistor Ra= (1/10 + 1/5 )-1 = 3,33 ohm |dw:1330119198577:dw| Ra and R2 (in series) can be replaced by a single resistor Rb=(3,33+R2) |dw:1330119280784:dw| Then Rb and R1 in parallel is equivalent to a single resistor Rc=(1/Rb + 1/R1)-1 = \[R1(10 +3R2) / (3(R1+R2) +10)\] Now that things have been simplified; the voltage across Rc is the same as that of our combined resistors into Rc (a) V(R1)= Rc*E/(Rc + 2)= \[(R1E(10+3R2))/(16R1+6R2+3R1R2+20)\] Coming back to our initial circuit: The volatge V across R1= R1*I (b) I= V/R1 = [(R1E(10+3R2))/(R1*(16R1+6R2+3R1R2+20))\] Numerical Answer: (a) V= (3*10*(10+3(8))/(16(3)+6(8)+3(3*8)+20) = 5,43 V (b) I= (3*10*(10+3(8))/3((16(3)+6(8)+3(3*8)+20))=1,81 A
A free spreadsheet program for Windows is in Open Office org's collection. It does most things quite well. Here is a better file to view for analyzing this circuit.
@Azagen You come up with valid answers. Yet your formulas must be coming from a equation calculator, as I can bearly figure our where they might even come from, and in one place you use -1 assumably for inversion or the power of negative 1.