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
A resistor with R = 8.09 ohm resistance is connected to a real (i.e. non-ideal) battery as shown in the figure.
The battery produces an electromotive force of ? = 16.9 V. When it is connected to the load resistor above, the terminal voltage of the battery drops down to ?Vt = 12.7 V.

A.) What is the internal resistance r of the battery?
B.) What is the electric current flowing in the circuit?
C.) What is the power dissipated by the external resistor?
D.) How much power is dissipated inside the battery?
E.) What is the efficiency of the circuit in percent?


Edit
Added Mon, 20 Jul '15
Community
1
Comment
Add a Comment
Solutions
0
a) If Voc=16.9 and Vload=12.7 with a load of 8.09 ohms, the current is 12.7/8.09=1.56 amp. This means the battery resistance is (16.9-12.7)/1.56=2.69 ohms

b) 1.56 amp

c) P=V*I=12.7*1.56=19.8 watts

d)1.56*1.56*2.69=6.54watt

e) eff=Pout/Pin=12.7/16.9=75%.
Edit
Added Mon, 20 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.