a. [tex]I = 6. 1[/tex] × [tex]10^-4[/tex] A
b. [tex]I = 2. 6[/tex] × [tex]10^-3[/tex] A
c. [tex]I = 0. 04[/tex] 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.
How to determine the current
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
[tex]I = \frac{12}{19700}[/tex]
[tex]I = 6. 1[/tex] × [tex]10^-4[/tex] A
B. V = 9V
R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series
[tex]I = \frac{12}{4700}[/tex]
[tex]I = 2. 6[/tex] × [tex]10^-3[/tex] A
C. V= 12V
1/R = [tex]\frac{1}{750} + \frac{1}{1200} + \frac{1}{950 }[/tex] = [tex]3. 22[/tex] × [tex]10^-3[/tex]
R = [tex]\frac{1}{3. 22 * 10 ^-3}[/tex] = 310. 56 Ω
[tex]I = \frac{12}{310. 56}[/tex]
[tex]I = 0. 04[/tex] 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:
https://brainly.com/question/14296509
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