Question
Download Solution PDFA capacitive accelerometer with linear displacement has a parallel plate configuration with distance between plates = d for a certain acceleration. If the plate mounting is such that inter- plate gap increases with acceleration, what can happen when the input acceleration is halved?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCapacitive Accelerometer
Definition: A capacitive accelerometer is a type of accelerometer that measures acceleration by detecting changes in capacitance due to the relative displacement of its plates. The system typically consists of a movable plate and a fixed plate, forming a parallel plate capacitor. The distance between these plates changes with acceleration, causing a corresponding change in capacitance.
Working Principle: The capacitance (C) of a parallel plate capacitor is given by:
C = ε × A / d
where:
- ε: Permittivity of the medium between the plates
- A: Area of the plates
- d: Distance between the plates
When the accelerometer experiences acceleration, the movable plate shifts, altering the distance d between the plates. This change in d modifies the capacitance, which is then measured to determine the corresponding acceleration.
Scenario: In the given problem, the accelerometer is configured such that the inter-plate gap increases with acceleration. This implies that when acceleration decreases (e.g., halved), the distance between the plates will also decrease. According to the capacitance formula, a decrease in d will increase the capacitance, as C is inversely proportional to d.
Correct Option Analysis:
The correct option is:
Option 4: The distance between plates is d/2 and capacitance doubled.
Explanation:
When the input acceleration is halved, the inter-plate gap decreases to half of its original value, i.e., the new distance between the plates becomes d/2. Substituting this into the capacitance formula:
Initial Capacitance, C₁ = ε × A / d
New Capacitance, C₂ = ε × A / (d/2) = 2 × (ε × A / d) = 2 × C₁
This shows that when the distance between the plates is halved, the capacitance doubles. Therefore, the correct answer is Option 4.
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