What is the approximate rate of change of resistance with...

What is the approximate rate of change of resistance with illumination for a photoconductive cell with the characteristics of Fig. 26 for the ranges (a) $0.1 \rightarrow 1 \mathrm{k} \Omega$, (b) $1 \rightarrow 10 \mathrm{k} \Omega$, and (c) $10 \rightarrow \mathrm{k} \Omega ?$ (Note that this is a log scale.) Which region has the greatest rate of change in resistance with illumination?

Key Concepts

Illumination-Dependent Resistance

Illumination-dependent resistance describes the phenomenon where the resistance of a material changes as a function of the light intensity it receives. In photoconductive materials, increased illumination typically leads to decreased resistance, which is a key characteristic leveraged in light sensing technologies.

Rate of Change in Resistance

The rate of change in resistance refers to how rapidly the resistance value shifts with variations in illumination. This concept involves determining the sensitivity of the photoconductive cell over different ranges, which is crucial for assessing performance in applications that require precise light detection.

Photoconductive Cells

Photoconductive cells, commonly known as photoresistors, are components whose electrical resistance varies inversely with the intensity of incident light. They are used in circuits to detect changes in illumination, converting light levels into measurable electrical signals.

Logarithmic Scale Analysis

Logarithmic scale analysis involves using a log scale to plot data, which is particularly useful when the variable spans several orders of magnitude. In the context of photoconductive cells, a log scale can linearize the relationship between resistance and illumination, making it easier to evaluate and compare the rate of change over various resistance ranges.

Transcript

00:01 Hi, here this circuit indicates a photoresistor, photoresistor whose resistance depends inversely on the sensitivity, intensity of light.

00:23 It is in series with a resistance of 1 .0 kilo ohm, a voltmeter standard voltmeter has been put in parallel with this 1 .0 kilo ohm and this is the battery providing 9 .0 volt to this circuit.

00:45 In the first part of the problem on the sunny day it is measured that the resistance of photoresistor that was 0 .56 kilo ohm and as these two are in series this r p and r these are in series.

01:12 So, net resistance equivalent resistance of the circuit that will be 0 .56 kilo ohm plus 1 kilo ohm means it comes out to be equal to 1 .56 kilo ohm or we can say this is 1560 ohm.

01:32 So, current in the circuit that will be given by i is equal to v by r using ohms law means battery voltage that is 9 volt net resistance 1560.

01:44 So, the reading of voltmeter as the current will remain same in series.

01:51 So, v is equal to i into r current that is 9 by 1560 resistance that is 1 kilo ohm or 1000 ohm.

02:01 So, the reading of voltmeter calculated to be equal to 5 .8 volt one of the answer for the first part of the problem.

02:11 Then on the cloudy day the light intensity will go down.

02:23 So, the resistance of photoresistor will increase it has become now 4 .0 kilo ohm.

02:31 So, now equivalent resistance that will be 4 .0 plus 1 .0 which was in series means this is 4 .5 .0 kilo ohm or we can say this is 5000 ohm.

02:48 So, current in the circuit again using ohms law 9 by 5000 volt...

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