Problem 1 - Applied Physics

Question

A cell has an emf of \(1.50 \mathrm{~V}\) and an internal resistance of \(0.0450 \Omega\). If there is \(0.250 \mathrm{~A}\) in the cell, what voltage is applied to the external circuit?

Step-by-Step Solution

Answer

The voltage applied to the external circuit is 1.48875 V.

Step 1: Identify the Given Variables

We have the emf (electromotive force) of the cell as \(1.50 \, \mathrm{V}\), the internal resistance of the cell is \(0.0450 \, \Omega\), and the current flowing through the circuit is \(0.250 \, \mathrm{A}\).

Step 2: Recall the Formula for Voltage Across External Circuit

The voltage across the external circuit \(V\) can be calculated using the formula: \( V = \text{emf} - I \times r \), where \(I\) is the current and \(r\) is the internal resistance of the cell.

Step 3: Calculate the Voltage Drop Across Internal Resistance

Substitute the current \(I = 0.250 \, \mathrm{A}\) and internal resistance \(r = 0.0450 \, \Omega\) into the formula for voltage drop: \( V_{\text{drop}} = I \times r = 0.250 \, \mathrm{A} \times 0.0450 \, \Omega = 0.01125 \, \mathrm{V} \).

Step 4: Determine the Voltage Applied to the External Circuit

Subtract the internal voltage drop from the emf to find the voltage across the external circuit: \( V = 1.50 \, \mathrm{V} - 0.01125 \, \mathrm{V} = 1.48875 \, \mathrm{V} \).

Key Concepts

Electromotive Force (EMF)

Electromotive Force, commonly abbreviated as EMF, describes the potential difference generated by a source such as a battery or a generator. It is the "force" that pushes the electric charge around a circuit. Think of it as the battery's potential to do work in an electrical circuit. The EMF is measured in volts (V) and symbolized by \( \varepsilon \).

In an ideal scenario, the EMF of a cell represents the maximum potential difference between its terminals when no current flows. So, if the EMF is stated as \(1.50\, \mathrm{V}\) like in the problem above, this would be the open-circuit voltage across the terminal when the cell is not connected to any load.

However, when a current flows through the circuit, the actual voltage delivered to the external circuit might be less due to internal factors. This is important as EMF is considered before internal resistances come into play.

Internal Resistance

Internal resistance is an intrinsic property of electrical power sources like cells or batteries that opposes the flow of current. Every cell has some amount of resistance within it, even though it might not be very noticeable.

This resistance is usually represented by \( r \) and results in energy being dissipated as heat when a current passes through the cell. The effect of this is a reduction in the voltage available to the external circuit. For the given exercise, the internal resistance of the cell is given as \(0.0450 \Omega\).

The greater the internal resistance, the more the voltage drop when current passes through. This is calculated by the formula \( V_{\text{drop}} = I \times r \), where \( I \) is the current and \( r \) is the internal resistance. This drop is why the actual voltage on an external circuit may be less than the EMF.

Current and Voltage Relationship

The relationship between current and voltage in a circuit can be best understood through Ohm's Law, which connects voltage (\(V\)), current (\(I\)), and resistance (\(R\)). According to this law, \( V = I \times R \).

For circuits involving a cell, the voltage across the external circuit can be viewed as the EMF minus the voltage drop across the internal resistance. This is formulated as \( V = \text{EMF} - I \times r \), which allows us to calculate the effective voltage on our external load.

In the example problem, you can observe how the current (\(0.250 \mathrm{~A}\)) and internal resistance (\(0.0450 \Omega\)) influence the final external voltage reading. Using the relationship, the problem clarifies how the total voltage is partly "used up" in overcoming the internal resistance, reducing the effective voltage supplied to the external circuit.