Answer:
a)Q=71.4 μ C
b)ΔV' = 10.2 V
Explanation:
Given that
C ₁= 8.7 μF
C₂ = 8.2 μF
C₃ = 4.1 μF
The potential difference of the battery, ΔV= 34 V
When connected in series
1/C = 1/C ₁ + 1/C₂ + 1/C₃
1/ C= 1/8.4 +1 / 8.4 + 1/4.2
C=2.1 μF
As we know that when capacitor are connected in series then they have same charge,Q
Q= C ΔV
Q= 2.1 x 34 μ C
Q=71.4 μ C
b)
As we know that when capacitor are connected in parallel then they have same voltage difference.
Q'= C' ΔV'
C'= C ₁+C₂+C₃ (For parallel connection)
C'= 8.4 + 8.4 + 4.2 μF
C'=21 μF
Q'= C' ΔV'
Q'=3 Q
3 x 71.4= 21 ΔV'
ΔV' = 10.2 V
Answer:
K = 373.13 N/m
Explanation:
The force of the spring is equals to:
Fe - m*g = 0 => Fe = m*g
Using Hook's law:
K*X = m*g Solving for K:
K = m/X * g
In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:
K = 10 / 0.0268 = 373.13N/m
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Answer:
The capacitance is 11 F for half and fully charged capacitor.
Explanation:
Capacitance of capacitor is given by the expression
Where ε is the vacuum permittivity, A is the area of plates and d is the separation between plates.
So capacitance does not depend upon charge and potential. So capacitance fully and half charged capacitors are same.
Here the capacitance is 11 F for half and fully charged capacitor.
