# Current Limiting resistor

Discussion in 'General Electronics Chat' started by cooded, Nov 10, 2010.

1. ### cooded Thread Starter Member

Jul 20, 2007
28
0
Hi,

I am a complete newbie to electronics. I am still learning the basics.I wanted to know how a current limiting resistor worked.If we add a current limiting resistor in series with a circuit then the current flowing through the resistor as well as the circuit connected after the resistor has to be same since they are connected in series.So how is the resistor limiting any current.Also there is going to be a potential drop across the resistor then the and hence the circuit will have lower voltage across it.However the same resistor connected in parallel makes more sense because it is going to divide the current and hence limit the current to the circuit and there wont be any potential drop too.
I know i am going wrong some where.Please help me out.
Thanks

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
I would refer you to our Ebook - http://www.allaboutcircuits.com/vol_1/chpt_5/1.html

Do recall Ohm's law, in the form I = E/R. Given that E in a series circuit remains the same, increasing R (adding that limiting resistor) means that I will be smaller (current is limited).

3. ### iONic AAC Fanatic!

Nov 16, 2007
1,422
68
You did mean to say that I in a series circuit remains the same? Yes?
Thus by adding a component in a series circuit you add a resistance that drops voltage, resulting in power consumption. This changes the current for all components in that series setup.

The idea that all components in a series circuit have the same current only holds true with the existing components in place. adding more resistive components reduces the current.

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
It would have been more explicitly clear to say that adding to the total R in a series circuit must necessarily decrease the current.

5. ### iONic AAC Fanatic!

Nov 16, 2007
1,422
68
Perhaps so, but with less learning.

6. ### Jaguarjoe Active Member

Apr 7, 2010
770
90
Lets say that you have a widget that operates at 5 volts and requires 1 amp of current. You power source is a 12 volt battery.
R=E/I, so 5 volts and 1 amp means your widget is equivalent to a 5 ohm resistance. Your source is 12 volts so you need to "drop" 7 of those volts to get down to 5 volts for the widget. Current is the same through a series circuit so to drop 7 volts you'll need a 7 ohm current limiting resistor (R=E/I = 7/1 = 7 ohms). Total resistance is 12 ohms for the circuit. I = E/R so the current flow is one amp and you'll also have the correct 5 volts for the widget. Now lets say you don't have a 7 ohm resistor and you use a 5 ohm one in its place. Your widget is 5 ohms + 5 for the resistor = 10 ohms total. Current now becomes I = E/R = 12v/10 ohms = 1.2 amps. The voltage across the widget now becomes E=IR = 1.2*5= 6 volts. Not a good thing for the widget.
Place any resistor in parallel with your widget and applying 12 volts to the circuit will result in death for the widget because there is nothing to limit the current to it because I= E/R = 12/5 = 2.4 amps. Voltage is the same across a parallel circuit so adding resistors in parallel does nothing for the widget.

Current limiting resistors are only useful for fixed current loads like relay coils and LEDs. For loads which vary in current a voltage regulator is required.

Mar 24, 2008
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