Evaluating Internal Resistance Examples

For the following circuit, calculate:

1. the potential differences \(V_\text{1}\), \(V_\text{2}\) and \(V_\text{3}\) across the resistors \(R_\text{1}\), \(R_\text{2}\), and \(R_\text{3}\).

2. the resistance of \(R_\text{3}\).

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

If the internal resistance is \(\text{0.1}\) \(\text{Ω}\), what is the emf of the battery and what power is dissipated by the internal resistance of the battery?

Step 1: 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.

Step 2: 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 \(R_\text{3}\).

Step 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*}

Step 4: 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*}

Step 5: 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*}

Step 6: 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*}

Step 7: 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.

\begin{align*} \mathcal{E}& = V+Ir\\ & = \text{23}+(\text{2})(\text{0.1})\\ & = \text{23.2}\text{ V} \end{align*}

Step 8: 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:

\begin{align*} P_r &= V_rI_r\\ & = (\text{0.2})(\text{2})\\ & = \text{0.4}\text{ W} \end{align*}

Step 9: Write the final answer

• \(V_\text{1} = \text{2.0}\text{ V}\)
• \(V_\text{2} = \text{6.0}\text{ V}\)
• \(V_\text{3} = \text{10.0}\text{ V}\)
• \(R_\text{3} = \text{7.5} \Omega\)
• \(\mathcal{E} = \text{23.2}\text{ V}\)
• \(P_r = \text{0.4}\text{ W}\)

The potential difference across a battery measures 18 V when it is connected to two parallel resistors of \(\text{4.00}\) \(\Omega\) and \(\text{12.00}\) \(\Omega\) respectively. Calculate the current through the cell and through each of the resistors. If the internal resistance of the battery is \(\text{0.375}\) \(\text{Ω}\) what is the emf of the battery?

Step 1: First draw the circuit before doing any calculations

Step 2: 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.

Step 3: 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*}

Step 4: 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*}

Step 5: 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*}

Step 6: 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:

\begin{align*}V &=I \cdot r \\ &=\text{6} \cdot \text{0.375} \\ &= \text{2.25}\text{ V}\end{align*}

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:

\begin{align*} \mathcal{E}& = V+Ir \\ & = \text{18}+\text{2.25} \\ & = \text{20.25}\text{ V} \end{align*}

Step 7: 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 \(\text{20.25}\) \(\text{V}\).