ToughSTEM
Sign Up
Log In
ToughSTEM
A question answer community on a mission
to share Solutions for all STEM major Problems.
Cant find a problem on ToughSTEM?
0
Three resistors with resistances of 5 ohm, 9 ohm, and 12 ohm, respectively, are connected in a series circuit with a 6.0 V battery. Find the potential difference across the 5 ohm resistor
Edit
Added Fri, 03 Jul '15
Community
1
Comment
Add a Comment
Solutions
0
if R1 and R2 are the resistances connected in series then effective resiatcne Rnet = R1+R2

if R1 and R2 are connected in parallel, then 1/Rnet =1/R1 +1/R2 or Rnet =   R1R2/(R1+R2)

same current passes through each of the ressitor in series combination and
same votage passes through eacf of the resistor in parallel combination

apply ohms law V = iR
where V = volatge and i = current

so here Rnet = 5+9+12 = 26 ohms

i = V/Rnet = 6/26 = 0.2307 Amps

so Vacross 5ohms = iR = 0.2307 * 5 = 1.153 Volts ------answer

other volatages are

V acrosss 9 ohms = 9*0.2307 = 2.0763 Volts

V across 12 ohms = 12* 0.2307 = 2.7684 Volts
Edit
Added Fri, 03 Jul '15
Community
1
Comment
Add a Comment
Add Your Solution!
Close

Click here to
Choose An Image
or
Get image from URL
GO
Close
Back
Add Image
Close
What URL would you like to link?
GO
α
β
γ
δ
ϵ
ε
η
ϑ
λ
μ
π
ρ
σ
τ
φ
ψ
ω
Γ
Δ
Θ
Λ
Π
Σ
Φ
Ω
Copied to Clipboard

Add Your Solution
Sign Up
to interact with the community. (That's part of how we ensure only good content gets on ToughSTEM)
OR
OR
ToughSTEM is completely free, and its staying that way. Students pay way too much already.
Almost done!
Please check your email to finish creating your account!
Welcome to the Club!
Choose a new Display Name
Only letters, numbers, spaces, dashes, and underscores, are allowed. Can not be blank.
Great! You're all set, .
A question answer community on a mission
to share Solutions for all STEM major Problems.
Why
The Purpose
How
The Community
Give Feedback
Tell us suggestions, ideas, and any bugs you find. Help make ToughSTEM even better.