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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?

Circuit

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  • $\begingroup$ 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. $\endgroup$ – march May 10 '19 at 3:28
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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 $I_T$.

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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:

circuit with equivalent resistance

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...

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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 $I_{\rm T} = \dfrac {V_{\rm T}}{R_1+(R_2||R_3)}$

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