Hello,
I have a simple circuit with single source providing 9V, R=3,9KΩ, two LEDs in parallel (red 1,8V and blue 2,7V).
What will happen if we turn on the voltage source?
Are my thoughts correct?
I is the same on each branch, I resistor is I LED1 + I LED2
We can't tell I total, because we don't know the resistance of each LED, but
I = U / R → 9V / 3.9k Ω = 0,0023 A
We have 0,0011A (1,1mA) on each LED, correct?
I think I am wrong, because forward voltage of LEDs are different (1,8 and 2,7V), that means they have different resistance, that means we have more current on one branch.
Voltage is the same on each branch, do we have three voltage drops (R and two LED) or only two (R and parallel LEDs)?
Which voltage do we have on each LED? To calculate that we need to get voltage drop of LED, subtract 9V and we get voltage drop for both LEDs
Voltage drop R
V=I*R → 2,3mA * 3.9k Ω = ~ 8,97V
Oh... that doesn't work because we need R total... wait... if I multiple by 5000 (R total would be greater then value of one resistor) I get 11V.
Another way is to assume the voltage drop of the resistor, between the resistor we set a voltage and calculate the current.
The current would be divided for both branches, if we have e.g. 23mA on resistor, 11mA on each LED.
But how can that help me, If I don't know the voltage drop of resistor.
I am lost here. I know that the red LED will glow but I don't know why and I don't know how to get the answer through calculation.
Would be amazing, if someone could help me out.
I have a simple circuit with single source providing 9V, R=3,9KΩ, two LEDs in parallel (red 1,8V and blue 2,7V).
What will happen if we turn on the voltage source?
Are my thoughts correct?
I is the same on each branch, I resistor is I LED1 + I LED2
We can't tell I total, because we don't know the resistance of each LED, but
I = U / R → 9V / 3.9k Ω = 0,0023 A
We have 0,0011A (1,1mA) on each LED, correct?
I think I am wrong, because forward voltage of LEDs are different (1,8 and 2,7V), that means they have different resistance, that means we have more current on one branch.
Voltage is the same on each branch, do we have three voltage drops (R and two LED) or only two (R and parallel LEDs)?
Which voltage do we have on each LED? To calculate that we need to get voltage drop of LED, subtract 9V and we get voltage drop for both LEDs
Voltage drop R
V=I*R → 2,3mA * 3.9k Ω = ~ 8,97V
Oh... that doesn't work because we need R total... wait... if I multiple by 5000 (R total would be greater then value of one resistor) I get 11V.
Another way is to assume the voltage drop of the resistor, between the resistor we set a voltage and calculate the current.
Voltage | Current |
1 | 0,00025 A or 0,25mA |
2 | 0,0005 |
3 | 0,0007 |
4 | 0,0010 |
5 | 0,0012 |
6 | 0,0015 |
7 | 0,0017 |
8 | 0,0020 |
9 | 0,0023 |
The current would be divided for both branches, if we have e.g. 23mA on resistor, 11mA on each LED.
But how can that help me, If I don't know the voltage drop of resistor.
I am lost here. I know that the red LED will glow but I don't know why and I don't know how to get the answer through calculation.
Would be amazing, if someone could help me out.
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