What would happen when two or more capacitors are connected in parallel across a potential difference?
1 answer:
Answer:
The Current Iₜ = I₁ + I₂ + I₃
Charge Qₜ = Q₁ + Q₂ + Q₃
Potential difference Vₜ = V₁ = V₂ = V₃
The total capacitance Cₜ = C₁ + C₂ + C₃
Explanation:
According to the attached image;
For parallel arrangements of capacitors, the current flowing through each of the capacitors sums up to the total current flowing through the circuit;
Iₜ = I₁ + I₂ + I₃
Also the charge storage by each capacitor sums up to give the total charge stored;
Qₜ = Q₁ + Q₂ + Q₃
The potential difference across each of the capacitors are the same and equal to the total voltage across the circuit;
Vₜ = V₁ = V₂ = V₃
The total capacitance equals the sum of the capacitances of each of the capacitors;
Cₜ = C₁ + C₂ + C₃
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Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U =
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid