A 30-volt battery is used to supply current to a 10-kiloohm resistor. What is the current through the resistor? What is the power dissipated by the resistor?
We can imagine this resistor connected in a simple series circuit with the 30-volt battery. We know that current will flow through this circuit. And for the first part of our question, we want to solve for the value of that current. We will call it capital 𝐼.
There is an equation that connects current, resistance, and voltage. It’s called Ohm’s law. And it says that the potential difference in a circuit, the voltage, is equal to the current in the circuit times the resistance of the circuit. In our circuit, the assumption is the only source of resistance is our 10-kiloohm resistor.
This means that when we plug in for our values of 𝑉 and 𝑅 in this equation to solve for current, the voltage 𝑉 is 30 volts while the resistance 𝑅 is simply the value of our resistor, 10 kiloohms or 10000 ohms. When we calculate this fraction, we find a result of 3.0 milliamps. That’s the current that runs through this circuit and, therefore, this resistor.
The next thing we want to solve for is the power dissipated by this resistor. Here’s the idea behind this power dissipation. As the current 𝐼 runs through this circuit and through the resistor, the resistor heats up and releases energy. That energy, since it’s released over some amount of time, equates to a power, a value in watts. This power dissipation happens just by virtue of the fact that this resistor, true to its name, resists the flow of current through it.
So what is that power dissipated by the resistor? We can recall another equation that’s helpful for electrical circuits. The power in our electrical circuit is equal to the current times the potential difference.
Note that we could use the equation in this form to solve for the power dissipated by the resistor or in several other forms. That’s because, for example, our potential difference 𝑉 could be replaced by 𝐼 times 𝑅 thanks to Ohm’s law. Or our current 𝐼 could be replaced by 𝑉 divided by 𝑅, again due to the relationship of Ohm’s law.
But since we know the current 𝐼 in the circuit, since we solved for it in our last step, and we’re given the potential difference 𝑉, this equation for power is as good as any. If we multiply our current 3.0 milliamps or 3.0 times 10 to the negative third amps by our potential 30 volts, we find a result of 90 milliwatts. That’s the power dissipated by this resistor.