# If r1 is in series with a parallel combination of r2, r3, and r4, when the resistance value of r2 increases, will the voltage across r3 increase?

Jun 24, 2014

The voltage drop on ${r}_{3}$ is increasing with an increase of ${r}_{2}$.

With increased resistance ${r}_{2}$, the combined resistance of parallel combination of ${r}_{2}$, ${r}_{3}$ and ${r}_{4}$ increases. This, in turn decreases the current going through the circuit according to the following calculations.

Let $R$ be a combined resistance of the parallel combination of ${r}_{2}$, ${r}_{3}$ and ${r}_{4}$.
Then, according to the laws for parallel circuits,
$\frac{1}{R} = \frac{1}{r} _ 2 + \frac{1}{r} _ 3 + \frac{1}{r} _ 4$
Obviously, $R$ is increasing with an increase of ${r}_{2}$.
The combined resistance of an entire circuit is $R + {r}_{1}$, and it's also increasing with an increase of ${r}_{2}$.
Therefore, the current in the circuit is decreasing since it's equal to
$I = \frac{V}{R + {r}_{1}}$,
where $V$ is the voltage of the source of electricity and $I$ is the current.

The voltage drop on the resistor ${r}_{1}$, which is equal to ${V}_{1} = I \cdot {r}_{1}$, becomes smaller with an increase of ${r}_{2}$ since the current is decreasing. Since combined drop of the voltage must be equal to $V$, the voltage drop on a parallel combination of ${r}_{2}$, ${r}_{3}$ and ${r}_{4}$ is increasing. This voltage drop is the same for all three resistors ${r}_{2}$, ${r}_{3}$ and ${r}_{4}$ since they are connected in parallel. So, the voltage drop on ${r}_{3}$ is increasing with an increase of ${r}_{2}$.

Quantitatively, the voltage drop on each and all parallel resistors (including ${r}_{3}$, of course) equals to
V_p = I*R = (V*R)/(R+r_1) = V/(1+r_1*1/R) = V/(1+r_1*(1/r_2 + 1/r_3 + 1/r_4)
From the above formula we see that increase of ${r}_{2}$ causes decrease of denominator and, hence an increase of ${V}_{p}$.