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. |
by Lav.J
December 06, 2022 |

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 |
by Foxx
December 06, 2022 |

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. |
by vanderghast
December 07, 2022 |

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 |
by Lav.J
December 08, 2022 |

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. |
by vanderghast
December 08, 2022 |

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