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:
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
Answer:
B. Electric potential
Explanation:
Answer:
Geographically Widespread
Answer:
480 mph
Explanation:
speed = 60 mph
time = 8 hours
distance = speed*time
= 60*8
= 480 mph