A resistor has a resistance of 1.0 times 10 to the two ohms and dissipates electrical energy at a rate of 2.0 watts. What is the current through the resistor? What is the voltage drop across the resistor?
We can call the resistor resistance, 1.0 times 10 to the two ohms, 𝑅. And we can name the power that the resistor dissipates, 2.0 watts, 𝑃. We want to solve for the current through the resistor, which we can call 𝐼, and the voltage drop that occurs across the resistor, which we’ll call 𝑉.
To solve for these values, let’s start by recalling two laws that relate electrical properties. First, let’s recall Ohm’s law. This law says that the voltage across a resistor is equal to the current through that resistor times the resistance value of the resistor 𝑅.
And second we can recall the equation for electric power. Electric power, 𝑃, is equal to current, 𝐼, times voltage 𝑉. In this exercise, we’re given two pieces of information: the resistance of the resistor and the power that it dissipates. We can rearrange Ohm’s law and the electrical power relationship to use these two given properties to solve for 𝐼 and 𝑉.
Since 𝑃 equals 𝐼 times 𝑉 and 𝑉 is equal to 𝐼 times 𝑅, we can substitute this expression for 𝑉 in our Ohm’s law equation into our power equation which gives 𝑃 equals 𝐼 squared times 𝑅. If we rearrange this equation to solve for current 𝐼, we see that it equals the square root of power, 𝑃, over resistance, 𝑅.
Since we’re given both of those values in the statement, we’re ready to plug in and solve for 𝐼. When we do and enter these values on our calculator, we see that, to two significant figures, 𝐼 is 0.14 amps. That’s the current running through the resistor. Now that we know the current, we can solve for the voltage across the resistor, 𝑉, by referring to our equation for electrical power, 𝑃 equals 𝐼𝑉.
Rearranging this equation to solve for 𝑉, we find that it equals power 𝑃 divided by current 𝐼 or 2.0 watts divided by 0.14 amps. Entering these values on our calculator, to two significant figures 𝑉, equals 14 volts. That’s the voltage drop across the resistor.