Electrical Power
Power and Energy: Electrical Power
A source of energy is required to drive current round a complete circuit. This is provided by batteries in the circuits you have been looking at. The batteries convert chemical potential energy into electrical energy. The energy is used to do work on the electrons in the circuit.
Power is a measure of how rapidly work is done. Power is the rate at which the work is done, work done per unit time. Work is measured in joules (J) and time in seconds (s) so power will be \(\frac{\text{J}}{\text{s}}\) which we call a watt (W).
In electric circuits, power is a function of both voltage and current and we talk about the power dissipated in a circuit element:
Definition: Electrical Power
Electrical power is the rate at which electrical energy is converted in an electric circuit. It calculated as:
\[P = I·V\]Power (P) is exactly equal to current (I) multiplied by voltage(V), there is no extra constant of proportionality. The unit ofmeasurement for power is the watt (abbreviated W).
Fact:
It was James Prescott Joule, not Georg Simon Ohm, who first discovered the mathematical relationship between power dissipation and current through a resistance. This discovery, published in 1841, followed the form of the equation: \(P = I^{2}R\) and is properly known as Joule's Law. However, these power equations are so commonly associated with the Ohm's Law equations relating voltage, current, and resistance that they are frequently credited to Ohm.
Equivalent forms
We can use Ohm's Law to show that \(P=VI\) is equivalent to \(P=I^2R\) and \(P=\frac{V^2}{R}\).
Using \(V=I\cdot R\) allows us to show: \begin{align*} P &= V\cdot I \\ & = (I \cdot R ) \cdot I\ \text{Ohm's Law} \\ & = I^2R \end{align*}
Using \(I=\frac{V}{R}\) allows us to show: \begin{align*} P &= V\cdot I \\ & = V \cdot \frac{V}{R}\ \text{Ohm's Law} \\ & = \frac{V^2}{R} \end{align*}
Example: Electrical Power
Question
Given a circuit component that has a voltage of \(\text{5}\) \(\text{V}\) and a resistance of \(\text{2}\) \(\text{Ω}\) what is the power dissipated?
Step 1: Write down what you are given and what you need to find
\begin{align*} V & = \text{5}\text{ V} \\ R & = \text{2}\text{ Ω} \\ P & = ? \end{align*}
Step 2: Write down an equation for power
The equation for power is: \[P = V^{2}R\]
Step 3: Solve the problem
\begin{align*} P & = \frac{V^{2}}{R} \\ & = \frac{(\text{5})^{2}}{(\text{2})} \\ & = \text{12.5}\text{ W} \end{align*} The power is \(\text{12.5}\) \(\text{W}\).
Example: Electrical Power
Question
Study the circuit diagram below:
The resistance of the resistor is \(\text{15}\) \(\text{Ω}\) and the current going through the resistor is \(\text{4}\) \(\text{A}\). What is the power for the resistor?
Step 1: Determine how to approach the problem
We are given the resistance of the resistor and the current passing through it and are asked to calculate the power. We can have verified that:\[P = I^2R\]
Step 2: Solve the problem
We can simply substitute the known values for \(R\) and \(I\) to solve for \(P\). \begin{align*} P &= I^2R \\ &= (\text{4})^2 \times \text{15} \\ &= \text{240}\text{ W} \end{align*}
Step 3: Write the final answer
The power for the resistor is \(\text{240}\) \(\text{W}\).
Example: Power in Series Circuit
Question
Two ohmic resistors (\(R_{1}\) and \(R_{2}\)) are connected in series with a cell. Find the resistance and power of \(R_{2}\), given that the current flowing through \(R_{1}\) and \(R_{2}\) is \(\text{0.25}\) \(\text{A}\) and that the voltage across the cell is \(\text{6}\) \(\text{V}\). \(R_{1}\) = \(\text{1}\) \(\text{Ω}\).
Step 1: Draw the circuit and fill in all known values.
Step 2: Determine how to approach the problem.
We can use Ohm's Law to find the total resistance R in the circuit, and then calculate the unknown resistance using:
\[R = R_{1} + R_{2}\]because it is in a series circuit.
Step 3: Find the total resistance
\begin{align*} V &=R\cdot I \\ R &= \frac{V}{I} \\ & =\frac{\text{6}}{\text{0.25}} \\ & = \text{24}\text{ Ω} \end{align*}
Step 4: Find the unknown resistance
We know that:
\[R = \text{24}\text{ Ω}\]and that
\[R_{1} = \text{1}\text{ Ω}\]Since
\[R = R_{1} + R_{2}\]\[R_{2} = R - R_{1}\]Therefore,
\[R = \text{23}\text{ Ω}\]Step 5: Solve the problem
Now that the resistance is known and the current, we can determine the power: \begin{align*} P&=I^2R \\ &=(\text{0.25})^2(\text{23}) \\ &=\text{1.44}\text{ W} \end{align*}
Step 6: Write the final answer
The power for the resistor \(R_2\) is \(\text{1.44}\) \(\text{W}\).
Tip:
Notice that we use the same circuits in examples as we extend our knowledge of electric circuits. This is to emphasise that you can always combine all of the principles you have learnt when dealing with any circuit.
Example: Power in Series and Parallel Networks of Resistors
Question
Given the following circuit:
The current leaving the battery is \(\text{1.07}\) \(\text{A}\), the total power dissipated in the circuit is \(\text{6.42}\) \(\text{W}\), the ratio of the total resistances of the two parallel networks \(R_{P1} : R_{P2}\) is 1:2, the ratio \(R_1 : R_2\) is 3:5 and \(R_3=\text{7}\text{ Ω}\).
Determine the:
- voltage of the battery,
- the power dissipated in \(R_{P1}\) and \(R_{P2}\), and
- the value of each resistor and the power dissipated in each of them.
Step 1: What is required?
In this question you are given various pieces of information and asked to determine the power dissipated in each resistor and each combination of resistors. Notice that the information given is mostly for the overall circuit. This is a clue that you should start with the overall circuit and work downwards to more specific circuit elements.
Step 2: Calculating the voltage of the battery
Firstly we focus on the battery. We are given the power for the overall circuit as well as the current leaving the battery. We know that the voltage across the terminals of the battery is the voltage across the circuit as a whole.
We can use the relationship \(P=VI\) for the entire circuit because the voltage is the same as the voltage across the terminals of the battery:\begin{align*}P &=VI \\V &= \frac{P}{I} \\&=\frac{\text{6.42}}{\text{1.07}} \\&= \text{6.00}\text{ V}\end{align*}
The voltage across the battery is \(\text{6.00}\) \(\text{V}\).
Step 3: Power dissipated in \(R_{P1}\) and \(R_{P2}\)
Remember that we are working from the overall circuit details down towards those for individual elements, this is opposite to how you treated this circuit earlier.
We can treat the parallel networks like the equivalent resistors so the circuit we are currently dealing with looks like:
We know that the current through the two circuit elements will be the same because it is a series circuit and that the resistance for the total circuit must be: \(R_T=R_{P1}+R_{P2}\). We can determine the total resistance from Ohm's Law for the circuit as a whole:\begin{align*}V_{battery}&=IR_T \\R_T &=\frac{V_{battery}}{I} \\&=\frac{\text{6.00}}{\text{1.07}}\\&=\text{5.61}\text{ Ω}\end{align*}
We know that the ratio between \(R_{P1} : R_{P2}\) is 1:2 which means that we know:\begin{align*}R_{P1} &= \frac{1}{2}R_{P2} \ \ \text{and} \\R_T &= R_{P1} + R_{P2} \\& = \frac{1}{2}R_{P2} + R_{P2} \\&=\frac{3}{2}R_{P2} \\(\text{5.61}) &=\frac{3}{2}R_{P2} \\R_{P2} &= \frac{2}{3}(\text{5.61}) \\R_{P2} &= \text{3.74}\text{ Ω}\end{align*}and therefore:\begin{align*}R_{P1} &= \frac{1}{2}R_{P2} \\&=\frac{1}{2}(3.74) \\&= \text{1.87}\text{ Ω}\end{align*}
Now that we know the total resistance of each of the parallel networks we can calculate the power dissipated in each:\begin{align*}P_{P1} &= I^2R_{P1} \\ &= (\text{1.07})^2(\text{1.87}) \\ &= \text{2.14}\text{ W}\end{align*}and\begin{align*}P_{P2} &= I^2R_{P2} \\ &= (\text{1.07})^2(\text{3.74}) \\ &= \text{4.28}\text{ W} \end{align*}
Step 4: Parallel network 1 calculations
Now we can begin to do the detailed calculation for the first set of parallel resistors.
We know that the ratio between \(R_{1} : R_{2}\) is 3:5 which means that we know \(R_{1}= \frac{3}{5}R_{2}\). We also know the total resistance for the two parallel resistors in this network is \(\text{1.87}\) \(\text{Ω}\). We can use the relationship between the values of the two resistors as well as the formula for the total resistance (\(\frac{1}{R_PT}=\frac{1}{R_1}+\frac{1}{R_2}\))to find the resistor values:\begin{align*}\frac{1}{R_{P1}}&=\frac{1}{R_1}+\frac{1}{R_2} \\\frac{1}{R_{P1}}&=\frac{5}{3R_2}+\frac{1}{R_2} \\\frac{1}{R_{P1}}&=\frac{1}{R_2}(\frac{5}{3}+1) \\\frac{1}{R_{P1}}&=\frac{1}{R_2}(\frac{5}{3}+\frac{3}{3}) \\\frac{1}{R_{P1}}&=\frac{1}{R_2}\frac{8}{3} \\R_2&=R_{P1}\frac{8}{3} \\&=(\text{1.87})\frac{8}{3} \\&=\text{4.99}\text{ Ω}\end{align*}We can also calculate \(R_{1}\):\begin{align*}R_{1}&= \frac{3}{5}R_{2} \\&= \frac{3}{5}(\text{4.99}) \\&= \text{2.99}\text{ Ω}\end{align*}
To determine the power we need the resistance which we have calculated and either the voltage or current. The two resistors are in parallel so the voltage across them is the same as well as the same as the voltage across the parallel network. We can use Ohm's Law to determine the voltage across the network of parallel resistors as we know the total resistance and we know the current:\begin{align*}V &= I R \\&=(\text{1.07})(\text{1.87}) \\&=\text{2.00}\text{ V}\end{align*}
We now have the information we need to determine the power through each resistor:\begin{align*}P_1&=\frac{V^2}{R_1} \\&=\frac{(\text{2.00})^2}{\text{2.99}} \\&=\text{1.34}\text{ W}\end{align*}\begin{align*}P_2&=\frac{V^2}{R_2} \\&=\frac{(\text{2.00})^2}{\text{4.99}} \\&=\text{0.80}\text{ W}\end{align*}
Step 5: Parallel network 2 calculations
Now we can begin to do the detailed calculation for the second set of parallel resistors.
We are given \(R_3=\text{7.00}\text{ Ω}\) and we know \(R_{P2}\) so we can calculate \(R_4\) from:\begin{align*}\frac{1}{R_{P2}} &= \frac{1}{R_3}+\frac{1}{R_4} \\\frac{1}{\text{3.74}} &= \frac{1}{\text{7.00}}+\frac{1}{R_4} \\R_4&=\text{8.03}\text{ Ω}\end{align*}
We can calculate the voltage across the second parallel network by subtracting the voltage of the first parallel network from the battery voltage, \(V_{P2} = \text{6.00}-\text{2.00}=\text{4.00}\text{ V}\).
We can now determine the power dissipated in each resistor:\begin{align*}P_3&=\frac{V^2}{R_3} \\&=\frac{(\text{4.00})^2}{\text{7.00}} \\&=\text{2.29}\text{ W}\end{align*}\begin{align*}P_4&=\frac{V^2}{R_2} \\&=\frac{(\text{4.00})^2}{\text{8.03}} \\&=\text{1.99}\text{ W}\end{align*}
This lesson is part of:
Electric Circuits