# Video: Calculating the Potential Difference Required to Produce a Particular Current

What is the voltage supplied to a 1.44-kW-power short circuit through a 0.100-ฮฉ resistance?

02:25

### Video Transcript

What is the voltage supplied to a 1.44-kilowatt-power short circuit through a 0.100-ohm resistance?

To understand this idea of a short circuit, letโs consider a simple series circuit, with a power supply at a voltage ๐ and a load with a resistance ๐. With the circuit constructed this way, current in it has no option but to travel this one single path through the loop. But itโs possible to change the design of this circuit so that current avoids this resistive load ๐. That is, the current moves to a much lower resistance path which effectively denies this load of any significant current.

One way to do that, one way to starve this resistor of current, would be to create a branch in the circuit which gives the current another option, another path to follow, that has much less resistance. Indeed with the resistance of this branch being negligibly small, virtually all of the current in the circuit will choose, so to speak, to follow that path while virtually none will pass through the resistive load.

So by adding in this low resistance option for the current to flow through, weโve created a short circuit. In this particular example, we have a short circuit which doesnโt have zero resistance. But it has a very small resistance value, 0.1 ohm.

Weโre told, moreover, that the short circuit is exposed to 1.44 kilowatts of power. Weโre looking then for a relationship between power, resistance, and the voltage we want to solve for. We can write that the power in an electrical circuit is equal to the current in the circuit times the potential difference. And we also remember Ohmโs law, which tells us that the potential difference in a circuit is equal to the product of current and resistance.

If we combine these two equations, then we can write that the power in an electrical circuit is equal to ๐ divided by ๐, which is effectively the current ๐ผ multiplied by that potential difference ๐. This simplifies to ๐ squared over ๐.

And we now have an equation in terms of the three variables of interest: voltage, resistance, and power. Since we want to solve for voltage, if we multiply both sides of this equation by ๐ and then take the square root of both sides, we find that ๐ is equal to the square root of ๐ times ๐, the square root of power times resistance. We plug in 1.44 times 10 to the third watts for the power and 0.100 ohms for the resistance.

This square root comes out to 12.0 volts. Thatโs the voltage supplied to this particular short circuit.