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Hello, I'm teaching myself about circuits, I don't understand why my calculations are different than that of the book. The goal is to actually find I4, but I can't seem to see why I'm getting a different value for I2.

Here are the book calculations:

R2+3||4 = R2 + (R3R4 / R3+R4) = 330Ohms + (330Ohms)(560Ohms)/890Ohms = 538Ohms

I2 = Vs/R2+3||4 = 5.0V/538Ohms = 9.29mA

Here are my calculations:

1/Rt = 1/560 + 1/538 Rt = 1/0.0036444 Rt = 274.39

It = Vs/Rt It = 5.0/274.389 = 18.2mA

I2 = (R1/R1+Rt)It I2 = (560/560+274.39)18.2mA = 12.21mA???

This is my first time posting, so if my formatting is wack let me know. Thank you for your time.

You need a ground connection somewhere to get any results at all. Put a ground at the negative terminal of V6 and see if that helps

Voltage (drop) is the same between two given points whatever path you take. So, being interested in I2, we have, between points A and C:

5 = I2 * ( R2 + R3 || R4 )

Solving for I2.

R1 is not involved, unless you are interested in the total current supplied by the voltage source, in which case you also have

5 = I1 * R1

Solving for R1.

And total current by the voltage source: I1 + I2

You main flaw in your computation is that you made I2 dependent of R1, which is not.

When using the voltage divider formula, instead of using the total resistance of the right-side branch, I used the total resistance of the circuit.

"I2 = (R1/R1+Rt)It I2 = (560/560+274.39)18.2mA = 12.21mA??? " Should have been this -> I2 = (560/560+538)18.2 = 9.29mA

Thanks for the replies though

R1 and R2 are not in series and so, a voltage divider notion does not apply.

Make R3 = R4 = 0, for simplicity. Rtotal = R1 R2 / ( R1 + R2) Itotal = 5 / Rt

all that is ok, but now:

I2 = (R1/(R1 + Rtotal)) Itotal is not.

Indeed, I2 = 5 / R2 (always with R3=R4=0) as well as I1 = 5 / R1

Your formula for I2 leads to $$I_2 = \frac{ 5 R_1} {R_1 + \frac{R_1 R_2}{R_1+R_2}} \frac{R_1+R_2}{R_1 R_2}$$

Which "simplifies" to $$ \frac{5 (R_1 + R_2) ^2 }{R_1 R_2 (R_1 + 2 R_2)} $$

(unless I made a mistake, which is possible), but clearly weird.

A voltage divider is valid for two resistors in series (not in parallel), with a negligible difference in the current through each of them. That is not the case at all, here.