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Sorry if this question has been answered here before, but I just can't seem to find any info on this.

A little background. I am currently doing an automotive mechatronics course and, while I know some things about circuits, my knowledge on this subject is very limited. I tried to my best effort to solve this on my own but i just can't get matching values.

Here's what i know:

The total current (It) is 150 mA

The voltage drop through the LED is 3 V

The current through R3 is 20 mA

The LED used in the drawing is just a representation (I couldn't erase the name, sorry)

If anyone can help me understand how to get the voltage drop through R1, R2, R3 aswell as the current, I will be most grateful

enter image description here

Here's what I tried: Ohm's law on first resistor to calculate voltage drop: it came as 24 V, which would be impossible since the source is only 9 V

Assuming that on the parallel part there was a voltage drop of 3 V: This would mean that across R1 and R2 the total voltage drop would be equal to 6 V. I could calculate V1, V2 and the current aswell. V1 = 4.8V, V2= 1.2V I= 30 mA. From here I can get to I=Id+I3, and I3 = U3/R3. Assuming that R3 is indeed 20 ohm, I3 would come as 0.15 A, which is 5x times larger than the input current.

So, my values don't check out and from the feedback I am receiving here, it's not a problem on my end

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  • \$\begingroup\$ after you draw the circuit click "save and insert" ... there is no reason to use a screenshot \$\endgroup\$
    – jsotola
    Commented Sep 22, 2020 at 2:22
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    \$\begingroup\$ @jsotola Sorry about that. I didn't knew that I could draw the circuit here \$\endgroup\$
    – José Capela
    Commented Sep 22, 2020 at 2:27
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    \$\begingroup\$ There is something wrong with the question. If the total current is 150 mA, Ohm's Law says there will be 24 volts across R1 and 6 volts across R2 which is clearly not possible with a 9 volt battery. 3 volts across the LED and 20 mA in R3 also doesn't work. \$\endgroup\$
    – Peter Bennett
    Commented Sep 22, 2020 at 2:57
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    \$\begingroup\$ Please show us your attempt at resolving the question before asking for help. For example what stranges values did you get and how did you derive them. Otherwise it just looks like you're trying to have someone do your homework for you. \$\endgroup\$
    – mhaselup
    Commented Sep 22, 2020 at 3:21
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    \$\begingroup\$ may be 1. The total current (It) is 150 mA 2. The voltage drop through the LED is 3 V 3. The current through R3 is 20 mA are three different scenarios. \$\endgroup\$
    – Curious KP
    Commented Sep 22, 2020 at 3:57

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I am not clear about the sense of your circuit.

If the forward voltage of the diode is really 3 volts.. you will never reach the nessesary voltage. It = 9 / 220V = 40.5 mA. Ur3 = 20 * 40.5 = 810 mV. This is below the the 3 volts. => not current trough the diode and no light!

You need a voltage over 33 volts to get the LED on. Your 150mA * (160 + 40) + 3 = 33 volt.

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  • \$\begingroup\$ Technically not my circuit. It was a last minute exercise on my automotive electricity class. I copied all the values from the board, but it might've escaped something. I will have to ask the professor if these values are correct \$\endgroup\$
    – José Capela
    Commented Sep 22, 2020 at 14:42

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