How much electrical power is needed to melt a wire? I saw a video in which someone melted a metal wire with electricity.
How much electrical power would this take?
 A: The electric power running in a wire is given by $P = V^2/R$, where $V$ is the voltage and $R$ is the resistance. The resistance is given by $R = \rho_e l/A$ where $\rho_e$ is the resistivity of the metal, $l$ is the length of the wire, and $A$ is the cross sectional area. So the power is given by $P=A  V^2/\rho_e l$. This is the amount of heat per unit time produced as a result of connecting this wire to a $V$-volt battery.
Now, if the wire is supposed to melt quickly, we can assume that all of this heat is used for heating and melting of the wire, and there is not enough time for it to dissipate the heat to the environment. The heat required to do so is the heat required to raise the temperature to $3695 K$ (Edit: this number was originally given in the question) plus the heat required for melting:
$$Q = m C \Delta T + m L_f,$$ 
where $m$ is the total mass, $C$ is the specific heat (heat required per unit mass of raising the temperature with one degree), $\Delta T$ is the change in the temperature, and $L_f$ is the latent heat of melting (heat per unit mass required to melt). The mass of a wire is the given by its mass density times its volume, which is length times cross section: $m=\rho_m l A.$
For the wire to melt we need $Pt>Q$, where $t$ is the total time. Note that we need to keep $t$ small so that the heat does not dissipate to the environment. 
Let us put everything together:
$$Pt>Q\implies \frac{A  V^2 t}{\rho_e l}>\rho_m l A(C \Delta T + L_f)\implies V>\sqrt{\frac{\rho_m\rho_e l^2(C \Delta T + L_f)}t}.$$
From the melting point being $3695 K$, I can guess that your metal is tungsten. For tungsten we have 
$$\rho_m=19250 kg/m^3,\quad \rho_e=52.8 n\Omega m,\quad C=134J/kg K,\quad and\quad  L_f=192000J/kg. $$
Let us assume the room temperature is $295 K$. Then $\Delta T=3400K.$ Plugging in everything, we have:
$$V>\frac{l}{\sqrt{t}}\sqrt{19250 kg/m^3\times52.8 n\Omega m \times(134J/kg K\times3400K+192000J/kg) }=\frac{25.6vol.s^{-1/2}.m^{-1} l}{\sqrt{t}}.$$
Assuming I haven't made an algebraic mistake, and assuming all of the generated heat goes into the wire and not the environment, in order for the wire to melt in $1$ second, you will need $\sim 26$ volts per meter length of the wire to melt it.
Realistically, a lot of the heat will be lost to the environment. In particular, the hot wire starts radiating light which one needs to take into account.
Update 1: Let us add the effect of radiation. The power of radiation is given by $\sigma T^4 S$, where $\sigma =  5.6703\times 10^{-8} W/m^2K^4$, and $S = 2\pi r l$ is the surface area of the wire. I'm going to make the assumption that all of the radiation takes place at the melting temperature to simplify the problem. In order to include this, we need the input electric power minus the output radiation power together satisfy $Pt>Q$. That is
$$
(\frac{A  V^2}{\rho_e l}-\sigma T^4 S)\,t >\rho_m l A(C \Delta T + L_f)\implies V>l\sqrt{\frac{2\rho_e\sigma T^4}{r}+\frac{\rho_m\rho_e(C \Delta T + L_f)}t},
$$
where $r$ is the radius of the wire (assuming cylindrical wire).
To melt a wire with $1$ mm radius, in one second, we need about $42$ volts per meter of the wire. This still seems low to me. Can anyone catch a mistake in my calculation?
Update 2: As JMac mentioned in the comment below, the resistivity of tungsten is highly temperature dependent. The value provided above is at $20^\circ$C. The value of $\rho_e$ near the melting temperature is about $20$ times larger. So the answer is somewhere between $3$ to $4.5$ ($\sqrt{20}\sim4.5$) times larger than calculated above.  
A: This answers your comment, not the question itself (which doesn't provide enough information to answer).
To melt something you first need to provide heat to it till it reaches the melting point. The formula for that is $H = mc\Delta T$, where $m$ is the mass, $c$ is a constant that depends on the material you're melting, and $\Delta T$ is the change in temperature. For example to bring ice from -50 celsius to 0 celsius (when it melts), $\Delta T$ is 50.
After bringing the metal to the melting point, you need to actually melt it. The formula for that is given by $H = mL_f$, where $m$ is the mass, and $L_f$ is a constant known as the latent heat of fusion. This constant measures the amount of energy required to melt the material, and depends on the material in question.
Once you have the amount of heat required, you need to match that to the heat generated by electricity. This is given by the formula $H = I^2R$, where $I$ is the current and $R$ is the electrical resistance of the metal. The electrical resistance itself is dependent on factors such as the temperature of the metal, the type of metal, and so on. Note the resistance depends on the temperature of the metal, which is changing as you heat it. This means that you'll need to use calculus. If you're only after the order of magnitude, you can treat the electrical resistance of the metal as a constant, and elementary algebra will give you the answer.
A: I am giving you a very rough answer here and will try to read other more elaborate answers later to learn!
I have done welding on small household jobs such as repairing the metal railing and what not! I used to have a 75 amp welding machine and it easily melts 1/8 inch thick electrodes! 
So roughly 110volts. 75amps= 8250watts. 
And it melts it so fast that you need to learn how to control the job otherwise you will make a hole in the base material!
