# Clarity on Some Basics

richs2k6 Member
Join Date: Nov 2010
Location: United States
Posts: 13
Clarity on Some Basics

So from what I understand voltage remains the same in a parallel circuit and current remains the same in a series circuit. If I am correct then the following would be true?

Incandescent Light Bulb
Volts = 4.9
Amps = .3
Resistance = 16.3
Watts = 1.47

Power Source
Volts = 5.0
Amps = 7.0
Watts = 35

If hooked up in a series I could use 23 bulbs (7 amps divided by .3 amps)?

I guess what I don't get is that if this is true I couldn't hook up 2 bulbs to the same adapter in a series because there is only 5 volts of power, correct? If I wanted to run them in a series I would need a 100v adapter rated at .3 amps?

PJmax Group Moderator
Join Date: Oct 2012
Location: Northern NJ - USA
Posts: 58,577
voltage remains the same in a parallel circuit and current remains the same in a series circuit
That is correct.

The bulb needs a certain voltage to light. You add the voltage up of each light in the series loop to determine the voltage needed to run the loop.

You would need to parallel the lamps. That would be 7A / .3 = 23 lamps.
The lamps in series would be 4.9v x 23 = 113v

richs2k6 Member
Join Date: Nov 2010
Location: United States
Posts: 13
I guess I'm confused on how the amps play a role. I was working through a book and the example was hooking up a 3 watt LED to a 9 volt battery. So it wouldn't burn the LED out you had to run a resistor in the series. That makes sense since you need to dissipate the 6 volts that aren't necessary. But, the LED only requires .02 amps and battery produces 500ma. Will the LED only draw the .02 amps it needs and ignore the rest? Does the resistor not have an effect on the amperage?

richs2k6 Member
Join Date: Nov 2010
Location: United States
Posts: 13

PJmax Group Moderator
Join Date: Oct 2012
Location: Northern NJ - USA
Posts: 58,577
Sure..... the resistor definitely has an effect on the amperage.

To figure the resistor value you need to know the following....
1) supply voltage (9v)
2) the diode's forward voltage (2v for red)
3) the diode's forward current. (.02 or 20ma.)

The resistor needed would be around 390 ohms.

You aren't dissipating 6 volts. You are limiting the circuit current flow to 20ma.

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