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.

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- May 15th 2011, 10:04 PMsilvercatsHow 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 AMBAdhi
Have you learned a theory called Kirchhoff's(not sure about the spellings) two laws?

- May 16th 2011, 05:55 AMsilvercats
- May 16th 2011, 07:11 PMBAdhi
Here you have to use the Kirchoff's first law (or the current law), which is on any joint of the circuit, .

From this you'll get the current flowing through the middle resistor as

now use the Kirchoff's Voltage law which is on any closed circuit,

add this equation to two closed circuits to find and - May 16th 2011, 07:32 PMsilvercats
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 PMBAdhi
- May 16th 2011, 08:47 PMsilvercats
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 PMBAdhi
you cannot simplify resistors that easily, refer Kirchhoff's circuit laws and your text books, I guess you've already been taught about this laws