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Answer:
1) Bigger sector = 168.8pi
1) Smaller sector = 56.3pi
2) Bigger sector = 529.9
3) Smaller sector = 176.6
Step-by-step explanation:
1) Bigger sector = ×pi×15×15
= 168.75pi
1) Smaller sector = ×pi×15×15
= ×pi×15×15
= 56.25pi
2) Bigger sector = ×3.14×15×15
= 529.875
2) Smaller sector = ×3.14×15×15
= ×3.14×15×15
= 176.625
a. ×
A
b. ×
A
c. A
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
<h3>How to determine the current</h3>The formula for finding current
I = V/R
Where v = voltage
R = resistance
A. V = 12V
R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series
×
A
B. V = 9V
R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series
×
A
C. V= 12V
1/R = =
×
R = = 310. 56 Ω
A
It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
Learn more about Ohms law here:
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