Electrical Engineering Asked by Yolo Gamer on December 29, 2021
I am new to circuits/electrical stuff
Correct me if i am wrong.
if i have an led (2.1V / 0.02A) and if i give it 0.02A it will only take 2.1V or if i give it 2.1V it will take 0.02A
Ohms law ($I=frac{V}{R}$).
If start with 9V i can use ohms law to get the correct power for the led because i know how many amps it requires
My question is:
If i started with 9V. Could i drop the voltage down for a component that takes 5V without knowing how many amps it uses?
Sub-question:
Wouldn’t a circuit like this drain a battery or cause the circuit to short out?
No need for a voltage dropper with two resistors. Just assume (a fairly accurate assumption I might add) that, the LED has 2.1 volts across it then, put a resistor in series with it to the 9 volt rail. The value of the resistor sets the current: -
$$I = dfrac{text{9 volts - 2.1 volts}}{R} = dfrac{6.9}{R} = text{20 mA}$$
Rearrange to find R (it equals 345 ohms).
Wouldn't a circuit like this drain a battery or cause the circuit to short out?
It's extremely sub-optimal.
If i started with 9V. Could i drop the voltage down for a component that takes 5V without knowing how many amps it uses?
You could (and waste a lot of power) or, just use a voltage regulator - that is what they are for.
Answered by Andy aka on December 29, 2021
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