Does increasing the resistance in a branch of a parallel circuit decrease the overall current? 
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?
 A: 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.
A: 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.
A: 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.
A: 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.
A: 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.
