Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

There are a variety of sensors which work on the basis of changing resistance du

ID: 1769813 • Letter: T

Question

There are a variety of sensors which work on the basis of changing resistance due to an external influence. For example, force-sensitive resistors (FSR) will change their resistance when a force is a exerted (when you squeeze Computers are good at reading out voltages, so these devices are often combined ith a fixed-value resistor to create a voltage divider circuit. Consider the circuit shown below where the fixed resistor, R, has a resistance of 20.0 k and the FSR has a resistance of 60.0 k when there is no force acting on the FSR. A voltage V=5.00 V is being applied to the two series resistors. v FSR What is the current flowing through the FSR when there is no force being applied to the FSR? Submit Answer Tries 0/10 What is the current flowing through the fixed resistor when there is no force being applied to the FSR? Submit Answer Tries 0/10 Consider the negative terminal of the voltage source to be at a potential of zero Volts. What is the electric potential (voltage) at the location A between the two resistors? Submit Answer Tries 0/10 Suppose that if a force of 1.10 N is applied to the FSR, its resistance decreases to 8.00 k2. What would be the new current through the fixed resistor, R? Submit Answer Tries 0/10 What would be the electric potential at Location A between the two resistors with the applied force of 1.10 N?

Explanation / Answer

a] Current in FSR when F = 0 N is:

i = 5/(60000 + 20000) = 62.5 x 10-6 A.

b] Since the two resistors are in series, the current flowing through both will be the same. Therefore, the current through the fixed resistor will be: i = 62.5 x 10-6 A.

c] Potential at A can be determined using Kirchoff's voltage law.

so, Potential at A will be:

VA = 5 - (62.5 x 10-6 )(60000) = 1.25 volts.

d] New force dependent resistance = 8000 ohms

therefore, the new current will be: i = 5/(8000+20000) = 1.786 x 10-4 A.

e] VA = 5 - (1.786 x 10-4 )(8000) = 3.57 volts.