Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height
of 30.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision
1 answer:
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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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