# How to solve this simple circuit ?

• May 15th 2011, 10:04 PM
silvercats
How to solve this simple circuit ?
http://oi51.tinypic.com/2nk2qtg.jpg

I was able to find I1:
simplified and got total resistance=10/3 and V=10(coz parallel)
by putting V=IR
I got 10=I(10/3)
I=3A

How to find the I2?I think my I1 is wrong too.
thanks for reasing.
• May 16th 2011, 03:32 AM
Have you learned a theory called Kirchhoff's(not sure about the spellings) two laws?
• May 16th 2011, 05:55 AM
silvercats
Quote:

Have you learned a theory called Kirchhoff's(not sure about the spellings) two laws?

have heard that name before.IDK,I have learned some laws and hard to remember what is what :/
• May 16th 2011, 07:11 PM
Here you have to use the Kirchoff's first law (or the current law), which is on any joint of the circuit, $\sigma I=0$.

From this you'll get the current flowing through the middle resistor as $I_1-I_2$

now use the Kirchoff's Voltage law which is on any closed circuit, $\sigma E=0$

add this equation to two closed circuits to find $I_1$ and $I_2$
• May 16th 2011, 07:32 PM
silvercats
I know that ΣE=ΣIR(that is the kirchhoff's law I guess?) thing.Did according to that and never got the answer in the book .I don't understand above thing too ups
• May 16th 2011, 08:33 PM
Quote:

Originally Posted by silvercats
I know that ΣE=ΣIR(that is the kirchhoff's law I guess?) thing.

if this is what you've used in your first post, it is not Kirchoff's law, it is Ohm's law. Kirchoff's law is a developed from the Ohm's law.

Can you describe in detail how you got I1
• May 16th 2011, 08:47 PM
silvercats
parallel 4Ω and 2Ω: (1/4)+(1/2)=1/R =4/3Ω
that 4/3Ω and left side's 2Ω are =4/3 + 2 =10/3Ω
because there are two parallel batteries ,10v and 10v =10 ( <<< this is wrong i think)

and

by putting V=IR
I got 10=I(10/3)
I=3A
• May 16th 2011, 09:12 PM