Why don't we need to consider \$R_4\$ and \$R_5\$ when why calculate \$I_1\$?

Question

My book said that \$I_1=I_T \times \frac{R_1}{R_1+R_2}\$,and \$I_2=I_T \times \frac{R_4}{R_4+R_5}\$

^{simulate this circuit – Schematic created using CircuitLab}

I have two questions about these formula

1.Why don't we need to consider \$R_4\$ and \$R_5\$ when why calculate \$I_1\$? i know \$R_1\$ and \$R_4\$ or \$R_2\$ and \$R_5\$ are not series connected ,nor parallel connected,but we didn't find the equivalent resistance of \$R_1\$ and \$R_4\$ or \$R_2\$ and \$R_5\$,we just ignore \$R_4\$ and \$R_5\$ when why calculate \$I_1\$,why?

2.Why is \$I_2=I_T \times \frac{R_4}{R_4+R_5}\$?why isn't it \$I_2=I_1 \times \frac{R_4}{R_4+R_5}\$?

Answer 1

The short answers is that the value of I_T already incorporates the values for all of the other resistors and the voltage source, so if you know that, you don't need to know anything but the resistor values in question: R1 & R2.

Circuit Theory

The first thing to do in these situations is rearrange the circuit to something that makes more sense to you. This is the equivalent circuit:

^{simulate this circuit – Schematic created using CircuitLab}

So you see that R1 || R2 is in series with R4 || R5 as well as R3. Things in series have the same current but split the voltage, and things in parallel have the same voltage but split the current. To further confuse you, they named the current going through R2 as I1, but I'll refer to currents and voltage by their associated component, so I_R2 is the current through R2, and I_R1||R2 is the total current through the parallel resistors R1 and R2.

Here's the known equations relating voltage, current, and resistance:

Substituting the appropriate values from equations 4, 5, and 6 you will come up with the following relationships:

Equivalent relationships can be made regarding R4 and R5.

Answer 2

The implication is that the I_T that was given isn't bogus and was calculated with the entire circuit taken into account, including R4 and R5.

2.Why is \$I_2=I_T \times \frac{R_4}{R_4+R_5}\$?why isn't it \$I_2=I_1 \times \frac{R_4}{R_4+R_5}\$?

Because the circuit doesn't split that way. I_1 is not the same as I_T, and R1 and R2 are connected at both ends. Remember that I_T splits up between R1 and R2 and then rejoins afterwards. It then splits up again between R4 and R5.

As far as R4 and R5 are concerned, R1||R2 is a single resistor, and as far as R1 and R2 are concerned, R4||R5 is a single resistor.

If you redraw the circuit like this, does this help?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

Answer to both questions: So actually if you wanted to find the total current I_T where I_1 can be calculated from, then you would need to know what each resistor value is. That’s because I_T is the V4/R_equivalent. Looking at the circuit we see that I_T=I_1+ I_R1. This is just Kirchhoff current law. Also the top node at top of circuit, I_1+I_R1=I_T which is the same as above since current can’t just disappear from a circuit. Because of this I_T=I_2+I_R4. Note that I_R3=I_T. Now using current division formulas, I_1, I_2, I_R1, and I_R4 can be determined. Current division formulas may look odd but can be derived to show that they in fact are true.