10.4 Evaluating internal resistance in circuits

Note

This is a very similar question to what you have seen earlier. This is to highlight the fact that the approach when dealing with internal resistance is built on all the same principles you have already been working with.

Determine how to approach the problem

We are given the potential difference across the cell and the current in the circuit, as well as the resistances of two of the three resistors. We can use Ohm's Law to calculate the potential difference across the known resistors. Since the resistors are in a series circuit the potential difference is \(V = V_\text{1} + V_\text{2} + V_\text{3}\) and we can calculate \(V_\text{3}\). Now we can use this information to find the potential difference across the unknown resistor \(R_\text{3}\).

Calculate potential difference across \(R_\text{1}\)

Using Ohm's Law: \begin{align*} R_\text{1} &= \frac{V_\text{1}}{I} \\ I \cdot R_\text{1} &= I \cdot \frac{V_\text{1}}{I} \\ V_\text{1} &= {I}\cdot{R_\text{1}}\\ &= 2 \cdot 1 \\ V_\text{1} &= \text{2}\text{ V} \end{align*}

Calculate potential difference across \(R_\text{2}\)

Again using Ohm's Law: \begin{align*} R_\text{2} &= \frac{V_\text{2}}{I} \\ I \cdot R_\text{2} &= I \cdot \frac{V_\text{2}}{I} \\ V_\text{2} &= {I}\cdot{R_\text{2}}\\ &= 2 \cdot 3 \\ V_\text{2} &= \text{6}\text{ V} \end{align*}

Calculate potential difference across \(R_\text{3}\)

Since the potential difference across all the resistors combined must be the same as the potential difference across the cell in a series circuit, we can find \(V_\text{3}\) using: \begin{align*} V &= V_\text{1} + V_\text{2} + V_\text{3}\\ V_\text{3} &= V - V_\text{1} - V_\text{2} \\ &= 23 - 2 - 6 \\ V_\text{3}&= \text{15}\text{ V} \end{align*}

Find the resistance of \(R_\text{3}\)

We know the potential difference across \(R_\text{3}\) and the current through it, so we can use Ohm's Law to calculate the value for the resistance: \begin{align*} R_\text{3} &= \frac{V_\text{3}}{I}\\ &= \frac{\text{15}}{\text{2}} \\ R_\text{3}&= \text{7,5}~\Omega \end{align*}

Potential difference across the internal resistance of the battery

The value of the emf can be calculated from the potential difference of the load and the potential difference across the internal resistance.

Power dissipated in the battery

We know that the power dissipated in a resistor is given by \(P=VI=I^2R=\dfrac{V^2}{R}\) and we know the current in the circuit, the internal resistance and the potential difference across it so we can use any form of the equation for power:

Write the final answer

• \(V_\text{1} = \text{2,0}\text{ V}\)
• \(V_\text{2} = \text{6,0}\text{ V}\)
• \(V_\text{3} = \text{10,0}\text{ V}\)
• \(R_\text{3} = \text{7,5} \Omega\)
• \(\mathcal{E} = \text{23,2}\text{ V}\)
• \(P_r = \text{0,4}\text{ W}\)

First draw the circuit before doing any calculations

Determine how to approach the problem

We need to determine the current through the cell and each of the parallel resistors. We have been given the potential difference across the cell and the resistances of the resistors, so we can use Ohm's Law to calculate the current.

Calculate the current through the cell

To calculate the current through the cell we first need to determine the equivalent resistance of the rest of the circuit. The resistors are in parallel and therefore: \begin{align*} \frac{\text{1}}{R} &= \frac{\text{1}}{R_\text{1}} + \frac{\text{1}}{R_\text{2}} \\ &= \frac{\text{1}}{\text{4}} + \frac{\text{1}}{\text{12}} \\ &= \frac{3+1}{\text{12}} \\ &= \frac{\text{4}}{\text{12}} \\ R &= \frac{\text{12}}{\text{4}} = \text{3,00} \ \Omega \end{align*} Now using Ohm's Law to find the current through the cell: \begin{align*} R &= \frac{V}{I} \\ I &= \frac{V}{R} \\ &= \frac{\text{18}}{\text{3}} \\ I &= \text{6,00}\text{ A} \end{align*}

Now determine the current through one of the parallel resistors

We know that for a purely parallel resistor configuration, the potential difference across the cell is the same as the potential difference across each of the parallel resistors. For this circuit: \begin{align*} V &= V_\text{1} = V_\text{2} = \text{18}\text{ V} \end{align*} Let's start with calculating the current through \(R_\text{1}\) using Ohm's Law: \begin{align*} R_\text{1} &= \frac{V_\text{1}}{I_\text{1}} \\ I_\text{1} &= \frac{V_\text{1}}{R_\text{1}} \\ &= \frac{\text{18}}{\text{4}} \\ I_\text{1} &= \text{4,50}\text{ A} \end{align*}

Calculate the current through the other parallel resistor

We can use Ohm's Law again to find the current in \(R_\text{2}\): \begin{align*} R_\text{2} &= \frac{V_\text{2}}{I_\text{2}} \\ I_\text{2} &= \frac{V_\text{2}}{R_\text{2}} \\ &= \frac{\text{18}}{\text{12}} \\ I_\text{2} &= \text{1,50}\text{ A} \end{align*} An alternative method of calculating \(I_\text{2}\) would have been to use the fact that the currents through each of the parallel resistors must add up to the total current through the cell: \begin{align*} I &= I_\text{1} + I_\text{2} \\ I_\text{2} &= I - I_\text{1} \\ &= 6 - 4.5 \\ I_\text{2} &= \text{1,5}\text{ A} \end{align*}

Determine the emf

This total current through the battery is the current through the internal resistance of the battery. Knowing the current and resistance allows us to use Ohm's law to determine the potential difference across the internal resistance and therefore the emf of the battery.

Using Ohm's law we can determine the potential difference across the internal resistance:

We know that the emf of the battery is the potential difference across the terminal summed with the potential difference across the internal resistance so:

Write the final answer

The current through the cell is \(\text{6,00}\) \(\text{A}\).

The current through the \(\text{4,00}\) \(\Omega\) resistor is \(\text{4,50}\) \(\text{A}\).

The current through the \(\text{12,00}\) \(\Omega\) resistor is \(\text{1,50}\) \(\text{A}\).

The emf of the battery is \(\text{20,25}\) \(\text{V}\).

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.

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 \(\text{6,00}\) \(\text{V}\).

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.

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 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 (\(\frac{\text{1}}{R_PT}=\frac{\text{1}}{R_\text{1}}+\frac{\text{1}}{R_\text{2}}\)) 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*}

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*}

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:

The power dissipated by the internal resistance of the battery is: