Three identical capacitors are connected in parallel to a battery. if a total charge of q flows from the battery, how much charg
1 answer:
Each capacitor carry the same charge 'q'.
Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:
= V₁ +V₂ +V₃
This voltage may also be calculated using capacitance and charge;
V = Q/ C
= V₁ +V₂ +V₃
Provided that the total charge is 'q', hence the total voltage can be expressed as:
= (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)
Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.
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Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W= where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W=
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules
Answer:
See the attached image and the explanation below
Explanation:
We must draw a schematic of the described problem, after the sketch it is necessary to make a free body diagram, at the time before and after cutting the cord.
These free body diagrams can be seen in the attached image.
First we perform a sum of forces on the x & y axes before cutting the cord, to be able to find the T tension of the wire. (This analysis can be seen in the attached image).
In this way we get the T-wire tension equation, before cutting.
Now we make another free body diagram, for the moment when the wire is cut (see in the attached diagram).
It is important to clarify that when the cord is cut, the system will no longer be in statically, therefore newton's second law will be used for summation of forces which will be equal to the product of mass by acceleration.
Finally with equations 1 and 2 we can find the K ratio.
