More Examples on Evaluating Internal Resistance
Evaluating Internal Resistance In Circuits: Applications
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 external circuit is \(\text{6.42}\) \(\text{W}\), the ratio of the total resistances of the two parallel networks \(R_{P\text{1}} : R_{P\text{2}}\) is 1:2, the ratio \(R_\text{1} : R_\text{2}\) is 3:5 and \(R_\text{3}=\text{7.00}\text{ Ω}\).
Determine the:
- potential difference of the battery,
- the power dissipated in \(R_{P\text{1}}\) and \(R_{P\text{2}}\), and
- if the battery is labelled as having an emf of \(\text{6.50}\) \(\text{V}\) what is the value of the resistance 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 potential difference 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 potential difference across the terminals of the battery is the potential difference across the circuit as a whole.
We can use the relationship \(P=VI\) for the entire circuit because the potential difference is the same as the potential difference 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 potential difference across the battery is \(\text{6.00}\) \(\text{V}\).
Step 3: Power dissipated in \(R_{P\text{1}}\) and \(R_{P\text{2}}\)
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: \(R_{Ext}=R_{P\text{1}}+R_{P\text{2}}\). We can determine the total resistance from Ohm's Law for the circuit as a whole:\begin{align*}V_{battery}&=IR_{Ext} \\R_{Ext} &=\frac{V_{battery}}{I} \\&=\frac{\text{6.00}}{\text{1.07}}\\&=\text{5.61}\text{ Ω}\end{align*}
We know that the ratio between \(R_{P\text{1}} : R_{P\text{2}}\) is 1:2 which means that we know:\begin{align*}R_{P\text{1}} &= \frac{\text{1}}{\text{2}}R_{P\text{2}} \ \ \text{and} \\R_T &= R_{P\text{1}} + R_{P\text{2}} \\& = \frac{\text{1}}{\text{2}}R_{P\text{2}} + R_{P\text{2}} \\&=\frac{\text{3}}{\text{2}}R_{P\text{2}} \\(\text{5.61}) &=\frac{\text{3}}{\text{2}}R_{P\text{2}} \\R_{P\text{2}} &= \frac{\text{2}}{\text{3}}(\text{5.61}) \\R_{P\text{2}} &= \text{3.74}\text{ Ω}\end{align*}and therefore:\begin{align*}R_{P\text{1}} &= \frac{\text{1}}{\text{2}}R_{P\text{2}} \\&=\frac{\text{1}}{\text{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_{P\text{1}} &= I^2R_{P\text{1}} \\ &= (\text{1.07})^2(\text{1.87}) \\ &= \text{2.14}\text{ W}\end{align*}and\begin{align*}P_{P\text{2}} &= I^2R_{P\text{2}} \\ &= (\text{1.07})^2(\text{3.74}) \\ &= \text{4.28}\text{ W} \end{align*} These values will add up to the original power value we had for the external circuit. If they didn't we would have made a calculation error.
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 \(R_{\text{1}} : R_{\text{2}}\) is 3:5 which means that we know \(R_{\text{1}}= \frac{\text{3}}{\text{5}}R_{\text{2}}\).We also know the total resistance for the two parallel resistors in this networkis \(\text{1.87}\) \(\text{Ω}\). We can use therelationship between the values of the two resistors as well as the formula for the totalresistance (\(\frac{\text{1}}{R_PT}=\frac{\text{1}}{R_\text{1}}+\frac{\text{1}}{R_\text{2}}\))to find the resistor values:\begin{align*}\frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{1}}+\frac{\text{1}}{R_\text{2}} \\\frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{5}}{3R_\text{2}}+\frac{\text{1}}{R_\text{2}} \\\frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{2}}(\frac{\text{5}}{\text{3}}+1) \\\frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{2}}(\frac{\text{5}}{\text{3}}+\frac{\text{3}}{\text{3}}) \\\frac{\text{1}}{R_{P\text{1}}}&=\frac{\text{1}}{R_\text{2}}\frac{\text{8}}{\text{3}} \\R_\text{2}&=R_{P\text{1}}\frac{\text{8}}{\text{3}} \\&=(\text{1.87})\frac{\text{8}}{\text{3}} \\&=\text{4.99}\text{ Ω}\end{align*}We can also calculate \(R_{\text{1}}\):\begin{align*}R_{\text{1}}&= \frac{\text{3}}{\text{5}}R_{\text{2}} \\&= \frac{\text{3}}{\text{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 potential difference or current. The two resistors are in parallel so the potential difference across them is the same as well as the same as the potential difference across the parallel network. We can use Ohm's Law to determine the potential difference 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_\text{1}&=\frac{V^2}{R_\text{1}} \\&=\frac{(\text{2.00})^2}{\text{2.99}} \\&=\text{1.34}\text{ W}\end{align*}\begin{align*}P_\text{2}&=\frac{V^2}{R_\text{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 \(R_\text{3}=\text{7.00}\text{ Ω}\) and we know \(R_{P\text{2}}\) so we can calculate \(R_\text{4}\) from:\begin{align*}\frac{\text{1}}{R_{P\text{2}}} &= \frac{\text{1}}{R_\text{3}}+\frac{\text{1}}{R_\text{4}} \\\frac{\text{1}}{\text{3.74}} &= \frac{\text{1}}{\text{7.00}}+\frac{\text{1}}{R_\text{4}} \\R_\text{4}&=\text{8.03}\text{ Ω}\end{align*}
We can calculate the potential difference across the second parallel network by subtracting the potential difference of the first parallel network from the battery potential difference, \(V_{P\text{2}} = \text{6.00}-\text{2.00}=\text{4.00}\text{ V}\).
We can now determine the power dissipated in each resistor:\begin{align*}P_\text{3}&=\frac{V^2}{R_\text{3}} \\&=\frac{(\text{4.00})^2}{\text{7.00}} \\&=\text{2.29}\text{ W}\end{align*}\begin{align*}P_\text{4}&=\frac{V^2}{R_\text{2}} \\&=\frac{(\text{4.00})^2}{\text{8.03}} \\&=\text{1.99}\text{ W}\end{align*}
Step 6: Internal resistance
We know that the emf of the battery is \(\text{6.5}\) \(\text{V}\) but that the potential difference measured across the terminals is only \(\text{6}\) \(\text{V}\). The difference is the potential difference across the internal resistance of the battery and we can use the known current and Ohm's law to determine the internal resistance:
\begin{align*}V&=I \cdot R \\R&=\frac{V}{I} \\& = \frac{\text{0.5}}{\text{1.07}} \\& = \text{0.4672897}\\& = \text{0.47}\text{ Ω}\end{align*}The power dissipated by the internal resistance of the battery is:
\begin{align*}P &=VI \\ & = \text{0.5}\cdot\text{1.07} \\ &=\text{0.535}\text{ W}\end{align*}Example: Internal Resistance and Headlamps (Exam Past Question)
Question
The headlamp and two IDENTICAL tail lamps of a scooter are connected in parallel to a battery with unknown internal resistance as shown in the simplified circuit diagram below. The headlamp has a resistance of \(\text{2.4}\) \(\text{Ω}\) and is controlled by switch \(\textbf{S}_1\). The tail lamps are controlled by switch \(\textbf{S}_2\). The resistance of the connecting wires may be ignored.The graph alongside shows the potential difference across the terminals of the battery before and after switch \(\textbf{S}_1\) is closed (whilst switch \(\textbf{S}_2\) is open). Switch \(\textbf{S}_1\) is closed at time \(\textbf{t}_1\).- Use the graph to determine the emf of the battery.(1 mark)
- WITH ONLY SWITCH \(\textbf{S}_1\) CLOSED, calculate the following:
- Current through the headlamp(3 marks)
- Internal resistance, \(r\), of the battery(3 marks)
- BOTH SWITCHES \(\textbf{S}_1\) AND \(\textbf{S}_2\) ARE NOW CLOSED. The battery delivers a current of \(\text{6}\) \(\text{A}\) during this period.Calculate the resistance of each tail lamp.(5 marks)
- How will the reading on the voltmeter be affected if the headlamp burns out? (Both switches \(\textbf{S}_1\) and \(\textbf{S}_2\) are still closed.)Write down only INCREASES, DECREASES or REMAINS THE SAME.Give an explanation.(3 marks)
Question 1
\(\text{12}\) \(\text{V}\)
(1 mark)
Question 2.1
Option 1:
\begin{align*} I & = \frac{V}{R} \\ & = \frac{\text{9.6}}{\text{2.4}} \\ & = \text{4 A} \end{align*}Option 2:
\begin{align*} \text{emf} & = IR + Ir \\ 12 & = I(\text{2.4}) + \text{2.4} \\ \therefore I & = \text{4 A} \end{align*}(3 marks)
Question 2.2
Option 1:
\begin{align*} \text{emf} & = IR +Ir \\ 12 & = \text{9.4} + 4r \\ r & = \text{0.6}\ \Omega \end{align*}Option 2:
\begin{align*} V_{lost} & = Ir \\ \text{2.4} & = \text{4}r \\ \therefore r & = \text{0.6}\ \Omega \end{align*}Option 3:
\begin{align*} \text{emf} & = I(R + r) \\ \text{12} & = \text{4}(\text{2.4} + r)\\ \therefore r & = \text{0.6}\ \Omega \end{align*}(3 marks)
Question 3
Option 1:
\begin{align*} \text{emf} & = IR +Ir \\ \text{12} & = \text{6}(R + \text{0.6}) \\ R_{\text{ext}} & = \text{1.4}\ \Omega \end{align*}\begin{align*} \frac{1}{R} & = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\ \frac{1}{\text{1.4}} & = \frac{1}{\text{2.4}} + \frac{1}{R} \\ R & = \text{3.36}\ \Omega \end{align*}Each tail lamp: \(R = \text{1.68}\ \Omega\)
Option 2:
\begin{align*} \text{Emf} & = V_{\text{terminal}} + Ir \\ 12 & = V_{\text{terminal}} + 6(\text{0.6}) \\ \therefore V_{\text{terminal}} & = \text{8.4}\text{ V} \end{align*}\begin{align*} I_{\text{2.4}\ \Omega} &= \frac{V}{R} \\ &= \frac{\text{8.4}}{\text{2.4}} \\ &= \text{3.5 A}\end{align*}\begin{align*} I_{\text{tail lamps}} & = 6 - \text{3.5} \\ & = \text{2.5}\text{ A} \\ R_{\text{tail lamps}} & = \frac{V}{I} \\ & = \frac{\text{8.4}}{\text{2.5}} \\ & = \text{3.36}\ \Omega \\ R_{\text{tail lamp}} & = \text{1.68}\ \Omega \end{align*}Option 3:
\begin{align*} V & = IR \\ \text{12} & = \text{6}(R)\\ R_{\text{ext}} & = 2\ \Omega \end{align*}\begin{align*} R_{\text{parallel}} & = 2 - \text{0.6} \\ & = \text{1.4}\ \Omega \\ \frac{1}{R} & = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\ \frac{1}{\text{1.4}} & = \frac{1}{\text{2.4}} + \frac{1}{R} \\ R & = \text{3.36}\ \Omega \end{align*}Each tail lamp: \(R = \text{1.68}\ \Omega\)
Option 4:
For parallel combination: \(I_{1} + I_{2} = 6\text{ A}\)
\begin{align*} \therefore \frac{V}{\text{2.4}} + \frac{V}{R_{\text{tail lamps}}} & = \text{6} \\ \text{8.4}\left(\frac{1}{\text{2.4}} + \frac{1}{R_{\text{tail lamps}}} \right) & = \text{6}\\ \therefore R_{\text{tail lamps}} & = \text{3.36}\ \Omega \\ R_{\text{tail lamp}} & = \text{1.68}\ \Omega \end{align*}(5 marks)
Question 4
Increases
The resistance increases and the current decreases. So \(Ir\) (lost volts) must decrease which leads to an increase in the voltage.
(3 marks)
[TOTAL: 15 marks]
This lesson is part of:
Electric Circuits