I was just reading about current divider in wikipedia http://en.wikipedia.org/wiki/Current_divider and applied the general formula mentioned there to this example http://www.allaboutcircuits.com/vol_1/chpt_6/3.html when I solve for Ir1 I get 5mA although the result in the textboox according to the general formula is 6 mA. What am I doing wrong? Thank you in advance
We are NOT mind readers! How can you POSSIBLY expect us to know how you got a particular wrong answer unless you show us HOW you went about coming up with that answer?
hello atomtm, welcome to the forum. as you have noticed by now, there are some expectations when it comes to the questions - you need to give us enough of information. the result of 5mA is incorrect for current through R1 when powered from 6V. this can be verified directly using Ohms law. I=V/R=6V/1k=6mA. but when using current divider, we normally do not know the voltage, we know current. in this example, total current is I=I1+I2+I3=6+2+3=11mA if we didn't know voltage but did know that total current is 11mA and resistor values are 1k 3k and 2k, we could still deduce individual currents through each resistor. to use current division rule we need to simplify circuit. if R1 is the branch of interest, we need to find equivalent resistance of rest of the circuit (all other branches). in this case that would be R2 and R3 which are in parallel so Rt= R2 || R3 this can be calculated by solving 1/Rt = 1/R2+ 1/R3 or by remembering simplification: Rt=R2*R3/(R2+R3) Rt=3*2/(3+2)=6/5 kOhm I=11mA, R1=1k, Rt=1.2 k; I1=I*Rt/(R1+Rt) I1=11mA * 1.2k /(1k +1.2k)= 11mA * 1.2/2.2 = 6mA it does not matter which method you use, if you don't make any mistake result will always be the same. if you wanted to apply this to find current through R2, then rest of the circuit is parallel connection of R1 and R3 Rt=R1*R3/(R1+R3)=1*2/(1+2)=2/3 kOhm I=11mA, R2=3k, Rt= (2/3) k; I2= I*Rt/(R2+Rt)=11mA * (2/3)/(3 + 2/3)=11mA * (2/3)/(9/3 + 2/3)= 11mA * (2/3) / (11/3)= 11mA * 2/11 = 2mA since we do know that V=6V we can do direct check, I2= 6V/R2= 6V/3k = 2mA
Well thank you all for your answers and I 'm really sorry for the dumb question i finally got the right answer!! Thank you again