With identical resistors, which fuse would blow first? The power supply voltage in this circuit is slowly increased. All of the resistors are identical as stated in the question. How do I know which fuse would blow first? Do I need to apply Kirchhoff's law? But even so, isn't the voltages dropped over each line are identical since the lines are in parallel? 
The correct answer is B.

 A: Short version
All of the current from the power source must go through either $\text{Fuse A}$ or $\text{Fuse B}$.  The effective resistance for $\text{Fuse A}$ is $2R$ while it's $\frac{1}{3}R$ for $\text{Fuse B}$.  Since $\text{Fuse B}$ has a much lower resistance, it gets most of the current.
The other fuses are all downstream of $\text{Fuse B}$, so they each only get part of $\text{Fuse B}$'s current.  So, $\text{Fuse B}$ would be the first to blow.
Long version
First, let's assume that each resistor has a resistance of $R$.  Let's suppose $R=1\Omega$.
Then, this circuit's basically 4 resistors in parallel.  The exception is that, in the second-to-top line, there're two resistors in series.  Since resistances in series are additive,
$$R_{\text{two in series}}=1\Omega+1\Omega=2\Omega.$$
The other three resistors in the circuit are in parallel, so their effective resistance is
$$R_{\text{three in parallel}}={\left(\left(1\Omega\right)^{-1}+\left(1\Omega\right)^{-1}+\left(1\Omega\right)^{-1}\right)}^{-1}=\frac{1}{3}\Omega.$$
Since $\text{Fuse A}$ and $\text{Fuse B}$ split the total current from the power source,
$$I_{\text{power source}}=I_{\text{A}}+I_{\text{B}}.$$
And since current through a branch is weighted by that branch's resistance, then $\text{Fuse A}$ gets
$$I_{\text{A}}=\left(1-\frac{\frac{1}{3}\Omega}{2\Omega+\frac{1}{3}\Omega}\right)I_{\text{power source}}=\frac{1}{7}I_{\text{power source}},$$
while $\text{Fuse B}$ gets
$$I_{\text{B}}=\left(1-\frac{2\Omega}{2\Omega+\frac{1}{3}\Omega}\right)I_{\text{power source}}=\frac{6}{7}I_{\text{power source}}.$$
Repeat for the final currents - $I_{\text{C}}$, $I_{\text{D}}$, and $I_{\text{E}}$ - and you should find that $I_{\text{B}}$'s the largest.  So, all else equal, that fuse should blow first.
If you don't care to do the math for the other three fuses, then you can instead reason that all of the current going through them must also go through $\text{Fuse B}$, so their currents have to be lower.
