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What is the explanation?

In the above question, why does R3 increase? If R2 increases, wouldn't the parallel combination's resistance increase? If so, wouldn't the circuit have less current? Then why would the voltage across R3 increase?

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  • $\begingroup$ Is there a source connected to this resistor network? If not, the answer is [c]. $\endgroup$ – The Photon Jul 31 '17 at 16:14
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When "the circuit has less current", then there will be a smaller voltage drop across R1. It's (a little bit) harder to calculate the currents in the individual branches R2-4, but you don't have to. If the sum of the voltage across R1 + R2-4 is constant (that is not explicitly stated, but I will assume you have a constant voltage source across the network), then lower current through R1 means lower voltage drop there, and more voltage across the parallel network.

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  • $\begingroup$ Why does having less current in the circuit only give a smaller voltage drop in R1 but not the parallel network? Why would the voltage across the parallel network increase? $\endgroup$ – paradox124 Jul 31 '17 at 15:09
  • $\begingroup$ The sum of the voltages across R1 and R2-4 must be constant (if the entire network is fed by a constant voltage source). $\endgroup$ – Floris Jul 31 '17 at 15:35
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The voltage drop across the parallel combination is equal to-

$ V - current*(R_1)$

where $V$ is the voltage across the terminals of the battery. This is so as the sum of the voltage drops across $R_1$ and the parallel combination is equal to $V$.

Hence if current decreases, voltage across the combination increases as $V$ and $R_1$ are constant.

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From an intuitive point of view, electricity will take the path of least resistance. When you increase the resistance on R2, current will be shunted away from this path to the other parallel branches. Current through R3 and R4 will increase, and since their resistance hasn't changed, the voltage increases as well.

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  • $\begingroup$ According to your answer what will happen to the voltage across $R_2$ ? However, when you increase $R_2$ the overall current in the circuit decreases. Then how will the current through $R_3$ increase, if the total current is decreasing ? $\endgroup$ – Mitchell Jul 31 '17 at 16:38
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By increasing $R_2$, you increase equivalent resistance of the three parallel resistors, say $R_{eq}$.

Consequently the current in the circuit decreases $I=\frac{V}{R_{1}+R_{eq}}$.

Before establishing relations to solve such type of questions always determine the parameter that remains constant. In this case, the voltage drop across the battery (or simply the battery's voltage) remains constant.

$V= constant$,

The voltage across $R_3$ will be,

$V_3=\frac{V}{R_{1}+R_{eq}}. R_{eq}$.

On simplifying,

$V_3=\frac{V}{\frac{R_1}{R_{eq}}+1}$,

From this relation you can see that on increasing the value of $R_2$ (or $R_{eq}$) the voltage drop across $R_3$ (or $R_2$ or $R_4$) increases.

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The voltage on all resistors in a parallel circuit is the same, so that means that the voltages on resistors $R_2,R_3,R_4$ in this example will always be the same. Since voltage is: $$U=I*R$$
this means that if we increase resistance value of element $R_2$, the voltage $U_2$ on that element will also increase, and since the voltages on all resistors in a parallel circuit are the same : $$U_2=U_3=U_4$$
that means that the voltage on $R_3$ will also increase.

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  • $\begingroup$ R1 is in series with R2, R3 and R4. Even if all the resistors were parallel the potential across each resistor would remain the same (equal to the batterys voltage) even when the resistance of any resistors is increased. Increasing R2 will lower the current, nothing will happen to the potential across the resistors. I apologize for being rude, but your answer is wrong. $\endgroup$ – Mitchell Jul 31 '17 at 16:51
  • $\begingroup$ @Mitchell I didn't see that R1 is in series with the others I'll edit that. How is it wrong? When you increase resistance the voltage also increases (unless you have constant voltage) and since R2 is parallel to R3 since voltage on R2 is increased voltage on R3 will increase as well. $\endgroup$ – Plexus Jul 31 '17 at 16:57
  • $\begingroup$ The battery is the same. Recall, when you calculate the current in a circuit you first calculate the equivalent resistance. When all the resistors are parallel, the potential drop across each resistor will be the same no matter what. But on increasing the resistance of a resistor the equivalent resistance will increase, which will only effect the current. The potential drop before increasing the resistance will be equal to the one after increasing (or decreasing) it. $\endgroup$ – Mitchell Jul 31 '17 at 17:14
  • $\begingroup$ @Mitchell I agree with what you said but it's only valid for a constant voltage and I don't see anywhere in the problem OP wrote that the voltage in this problem is constant. $\endgroup$ – Plexus Jul 31 '17 at 17:20
  • $\begingroup$ In such questions, voltage is always assumed to be constant (unless stated). $\endgroup$ – Mitchell Jul 31 '17 at 17:23

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