# Current Divider rule

If you have two resistors $$R_{A},R_{B}$$ connected in parallel with a battery, you can find the current that flows through one resistor ex.$$R_{A}$$ using the current divider rule:

$$I_{A}=\frac{V_{T}}{R_{A}}$$ Hence: $$I_{A}=\frac{I_{T}R_{B}}{R_{A}+{R_{B}}}$$

As far as I understand, this rule holds because the voltage drop in each resistor is equal to the total voltage of the source $$V_{T}$$, but why do I find books that apply it even if it is not? Take for example this circuit: Why does the rule apply for finding the current that passes through $$R_{2}$$ for example, even though its voltage drop isn't equal to the voltage of the battery?

• If your book says that the voltage across resistors 2 and 3 is the same as the voltage supplied by the battery, then your book is wrong. It is still true, however, that resistors 2 and 3 have the same potential difference (voltage) across them, because they are in parallel; this voltage will just end up being smaller than 22 V. – march May 10 '19 at 3:28

In this case it may help if you break it down in steps a little bit. So first replace $$R_2$$ and $$R_3$$ by their equivalent resistance ($$1/R_\mathrm{equiv} = 1/R_2 + 1/R_3$$), then calculate the total resistance and use that to get the current $$I_T$$.

This applies because the rule is referring to the voltage across the parallel resistors $$R_2$$ and $$R_3$$, which are represented as $$R_A$$ and $$R_B$$ in the equation $$I_A = \frac{I_T R_B}{R_A+R_B}$$.

The problem here is that the $$I_A$$ in the equation refers to the current through the resistance $$R_A$$, but there is also a point $$A$$ in the diagram. The current at point $$A$$ would normally also be written as $$I_A$$, but this is not the $$I_A$$ in the equation.

$$V_A = V_B$$ hence by Ohms Law, $$I_AR_A=I_BR_B$$, which you already have as your first equation, $$I_A = \frac{V_T}{R_T}$$, since the circuit can be redrawn as a circuit with equivalent resistance like so:

Now, since current is the same in all parts of a series circuit, $$I_T=I_A$$, and as the parallel resistance comes after the point $$A$$, we can use $$I_T$$ interchangeably with $$I_A$$, hence $$I_T=I_A=\frac{V_T}{R_T}$$.

The equivalent resistance $$R_{23}$$ can be found using the parallel resistance formula $$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...$$, here as $$\frac{1}{R_{23}}=\frac{1}{R_2}+\frac{1}{R_3}$$, or the shortcut when there are only two resistors, $$R_T = \frac{R_A R_B}{R_A + R_B}$$.

Notice the similarity to the equation $$I_A = \frac{I_T R_B}{R_A+R_B}$$?

You can then apply Ohms Law again again, substituting $$\frac{V_A}{R_A} = \frac{\frac{V_T}{R_T} R_B}{R_A+R_B}$$, and work your way through it to find the current through one resistor from the voltage drop and the resistor's contribution to the total resistance...

From your original equations with the subscripts changed as follows $$V_{\rm T} \rightarrow V_{\rm A},\, R_{\rm A} \rightarrow R_2$$ and $$R_{\rm B} \rightarrow R_3$$ for the circuit shown

$$I_2=\dfrac{V_{\rm A}}{R_2}$$ and $$I_2=\dfrac{I_{\rm T}R_3}{R_2+R_3}$$ where $$V_{\rm A} = V_{\rm T}- I_{\rm T} R_1$$ and $$I_{\rm T} = \dfrac {V_{\rm T}}{R_1+(R_2||R_3)}$$