0
\$\begingroup\$

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}\$

schematic

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}\$?

\$\endgroup\$
4
  • 2
    \$\begingroup\$ R1 and R2 are in parallel, and R4 and R5 are in parallel. R1//R2, is in series with R4//R5 \$\endgroup\$
    – Chu
    Apr 21, 2020 at 14:22
  • 1
    \$\begingroup\$ If I told you that the current that flows into the node below R1 and R2, flows through the parallel combination of R1 and R2 and then out the node at the top before splitting between R4 and R5 would that help? \$\endgroup\$
    – Buck8pe
    Apr 21, 2020 at 14:27
  • 1
    \$\begingroup\$ Your book is correct. \$\endgroup\$
    – Andy aka
    Apr 21, 2020 at 14:33
  • \$\begingroup\$ You do not need to include the R4 and R5 because you have already included them when you solve for I_T current. So, you already know the I_T current. Thus you can find I1 and I2 without including R4 and R5. \$\endgroup\$
    – G36
    Apr 21, 2020 at 14:43

3 Answers 3

2
\$\begingroup\$

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:

schematic

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: equations 1

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

equations 2

Equivalent relationships can be made regarding R4 and R5.

\$\endgroup\$
1
\$\begingroup\$

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?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
0
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.