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
<u>Answer</u>
3.7 Km south
<u>Explanation</u>
The definition of displacement is the distance traveled in a specific direction. It is the vector quantity. We add displacements like the way we add vectors.
Taking the direction towards North to be positive (+1.7 Km), the distance towards south would be negative (-5.4 Km).
Now lets add the two values.
(+1.7) + (-5.4) = 1.7 - 5.4
= - 3.7 Km But negative was towards south.
∴ Answer = 3.7 Km south.
Answer:
Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.
Explanation:
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Answer:
Explanation:
= Strain = 0.49
= 3.1 MPa
At t = Time = 32 s 
= 0.41 MPa
= Time-independent constant
Stress relation with time
at t = 32 s
The time independent constant is 16.0787 s
At t = 6
From the first equation
Answer:
Potential difference and charge will also increase.
Explanation:
Asking that :
What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?
The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.
And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:
Q = CV
Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.
Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.
