Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
chansharp
Group Title
Consider the circuit shown in the figure below. (Let R1 = 3.00 Ω, R2 = 8.00 Ω, and = 10.0 V.)
(a) Find the voltage across R1.
(b) Find the current in R1.
 2 years ago
 2 years ago
chansharp Group Title
Consider the circuit shown in the figure below. (Let R1 = 3.00 Ω, R2 = 8.00 Ω, and = 10.0 V.) (a) Find the voltage across R1. (b) Find the current in R1.
 2 years ago
 2 years ago

This Question is Closed

Tweedle_Dee Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

estebananaya Group TitleBest ResponseYou've already chosen the best response.0
i think you could first calculate the total resistance: Req = 2+ ((105)+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
 2 years ago

abdulnaz92 Group TitleBest ResponseYou've already chosen the best response.0
see thus image
 2 years ago

gdmellott Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Azagen Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

gdmellott Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

gdmellott Group TitleBest ResponseYou've already chosen the best response.0
@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.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.