Please answer !!!A company wants to install a sensor to monitor the light level in the offices of their buildings. The sensor contains an LDR which has a resistance of 10kohlms in daylight and 100kohlms in the dark . The choice of resistor in the circuit is between one of 25kohlms and 1 megaohlm. The input p.d to the sensing circuit of 12 v . State and explain which resistor is the best choice for the circuit in light and dark which each of the resistors . You could put your calculations in a table .

Answers

Answer 1

Answer:

Answer:

Explanation:

The sensor contains an LDR which has a resistance of 10kohlms in daylight and 100kohlms in the dark.

If the resistor in the circuit is 1 megaohlm, the total resistance in daylight and darkness will be 1.01 megaohms and 1.1 megaohlms.

The percentage difference = (1.1-1.01)/1.1*100% = 8.18%

If the resistor in the circuit is 25 kohlm, the total resistance in daylight and darkness will be 35 kohms and 125 kohlms.

The percentage difference = (125-35)/125*100% = 72%

With the input p.d to the sensing circuit fixed at 12 v, the sensing current will change according to the total resistance. A 72% difference is much more detectable. So the 25 kohm resistor is the better choice.

Answer 2

Answer:

Answer:

Explanation:

V=IR

I=12/(R of resistor + R of LDR)

R of LDR = 10kohm in light and = 100kohm in dark

R1 = 25kohm

R2 = 1Mohm

solve 4 current

light dark

R1 12/(25+10)=0.343mA 12/(25+100)=0.096mA

R2 12/(1000+10)=0.012mA 12/(1000+100)=0.011mA

so R1 is better as its easier 2 tell its light or dark

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Answers

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