Answer to Question #94498 in Electric Circuits for Sophie

Question

In class, I "drew a circuit" with a pencil. When the LED was moderately bright about 9mA was flowing and so (since I was using a 9V battery), the resistance was about 1 kΩ.

At this point, the distance

Answer 1

From ohm's law,

We know that

$R=\frac{V}{I}$

Or,

$V=RI=10^3\times9\times10^{-3}=9\ V\\So\ potential\ drop across\ LED\ will \ be\ 9\ V$

Power deliverd by LED will be $VI=9\times9\times10^{-3}=8.1\times10^{-2}\ W$

So at this time work done by LED will be $8.1\times10^{-2} \ J\ per\ second$