Simple Diode Circuit

Discussion in 'Homework Help' started by marineoceania, Dec 1, 2008.

1. marineoceania Thread Starter New Member

Dec 1, 2008
2
0
hey everybody, I have a simple question for you guys. On my homework I need to find the current flowing in the circuit.

The battery = 5 volts

R1 = 250 Ω
R2 = 180 Ω

I know the first (top left) corner is 5 volts and the bottom two are 0 volts (?). How do I find the current and the corner before the diode?

Thanks all!

2. mik3 Senior Member

Feb 4, 2008
4,846
70
If you assume the diode has a voltage drop of 0.7V then the current through th circuit is:

I=(5-0.7)/(R1+R2)

3. marineoceania Thread Starter New Member

Dec 1, 2008
2
0
hmm makes sense, but why would you add both resistors? Just curious haha

4. beenthere Retired Moderator

Apr 20, 2004
15,808
295
It's a series circuit.

5. leftyretro Active Member

Nov 25, 2008
394
5
Because there is only one current path and it has to pass through both resistors. The current doesn't know how many resistors there are, 1 or many, it just feels the total series resistance. One of the Kirchhoff's circuit laws covers the topic.