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?

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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?

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Answer 1 · 2014-06-24T04:31:26

The voltage drop on $r_{3}$ $r_3$ is increasing with an increase of $r_{2}$ $r_2$ .

With increased resistance $r_{2}$ $r_2$ , the combined resistance of parallel combination of $r_{2}$ $r_2$ , $r_{3}$ $r_3$ and $r_{4}$ $r_4$ increases. This, in turn decreases the current going through the circuit according to the following calculations.

Let $R$ $R$ be a combined resistance of the parallel combination of $r_{2}$ $r_2$ , $r_{3}$ $r_3$ and $r_{4}$ $r_4$ .
Then, according to the laws for parallel circuits,
$\frac{1}{R} = \frac{1}{r_{2}} + \frac{1}{r_{3}} + \frac{1}{r_{4}}$ $1/R = 1/r_2 + 1/r_3 + 1/r_4$
Obviously, $R$ $R$ is increasing with an increase of $r_{2}$ $r_2$ .
The combined resistance of an entire circuit is $R + r_{1}$ $R+r_1$ , and it's also increasing with an increase of $r_{2}$ $r_2$ .
Therefore, the current in the circuit is decreasing since it's equal to
$I = \frac{V}{R + r_{1}}$ $I = V/(R+r_1)$ ,
where $V$ $V$ is the voltage of the source of electricity and $I$ $I$ is the current.

The voltage drop on the resistor $r_{1}$ $r_1$ , which is equal to $V_{1} = I \cdot r_{1}$ $V_1=I*r_1$ , becomes smaller with an increase of $r_{2}$ $r_2$ since the current is decreasing. Since combined drop of the voltage must be equal to $V$ $V$ , the voltage drop on a parallel combination of $r_{2}$ $r_2$ , $r_{3}$ $r_3$ and $r_{4}$ $r_4$ is increasing. This voltage drop is the same for all three resistors $r_{2}$ $r_2$ , $r_{3}$ $r_3$ and $r_{4}$ $r_4$ since they are connected in parallel. So, the voltage drop on $r_{3}$ $r_3$ is increasing with an increase of $r_{2}$ $r_2$ .

Quantitatively, the voltage drop on each and all parallel resistors (including $r_{3}$ $r_3$ , of course) equals to
$V_{p} = I \cdot R = \frac{V \cdot R}{R + r_{1}} = \frac{V}{1 + r_{1} \cdot \frac{1}{R}} = \frac{V}{1 + r_{1} \cdot (\frac{1}{r_{2}} + \frac{1}{r_{3}} + \frac{1}{r_{4}})}$ $V_p = I*R = (V*R)/(R+r_1) = V/(1+r_1*1/R) = V/(1+r_1*(1/r_2 + 1/r_3 + 1/r_4)$
From the above formula we see that increase of $r_{2}$ $r_2$ causes decrease of denominator and, hence an increase of $V_{p}$ $V_p$ .

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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?

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