Problem 3 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 1.5-V dry cell can be tested by connecting it to a lowresistance ammeter. It should be able to supply at least 25 A. What is the internal resistance of the cell in this case, assuming it is much greater than that of the ammeter?

Step-by-Step Solution

Answer

The internal resistance of the cell is 0.06 ohms.

Step 1: Understanding the Problem

We are given a 1.5-V dry cell connected to an ammeter. The cell should supply at least 25 A. We need to find the internal resistance of the cell, assuming it is much greater than the ammeter's resistance.

Step 2: Applying Ohm's Law

Ohm's Law states that \( V = IR \), where \( V \) is voltage, \( I \) is the current, and \( R \) is the resistance. We know the voltage \( V = 1.5 \) volts and the current \( I = 25 \) amperes.

Step 3: Setting Up the Equation

We are to find the internal resistance \( R \) of the battery. Since the internal resistance is much greater than the ammeter's resistance, we approximate the total resistance as just the internal resistance of the cell. Replacing the values in Ohm’s Law gives us the equation: \( 1.5 = 25 \times R \).

Step 4: Solving for Internal Resistance

Rearrange the equation \( 1.5 = 25R \) to solve for \( R \). Divide both sides by 25, leading to: \( R = \frac{1.5}{25} \).

Step 5: Calculating the Final Answer

Perform the calculation: \( R = \frac{1.5}{25} = 0.06 \) ohms. Thus, the internal resistance of the cell is 0.06 ohms.

Key Concepts

Internal Resistance

Internal resistance is a concept that refers to the opposition within an electronic component, like a battery, to the flow of electric current. This is intrinsic to the component itself, meaning it's not caused by any external wires or devices connected to it. When current flows through a cell with internal resistance, there is a loss of voltage internally due to this resistance.

For practical understanding:

• Internal resistance causes the battery to lose some energy as heat, reducing the amount of energy available to the external circuit.
• Higher internal resistance means more voltage drop and less power output from the battery.
• It's a crucial factor when determining how well a cell can provide current to a load.

In exercises like the one with the 1.5-V dry cell, knowing the internal resistance helps us understand why a battery may underperform or what kind of load it can efficiently support.

Current Calculation

Calculating current is often a key task in electronics problems, grounded on the principles of Ohm's Law. Current, denoted as "I," is the flow of electric charge, measured in amperes (A). Ohm's Law establishes a relationship between current, voltage, and resistance, summed up as: \( V = IR \).

To calculate current:

• Identify the voltage across the circuit, provided in volts (V).
• Determine the total resistance in the circuit, given in ohms (\( \Omega \)).
• Use Ohm’s Law to rearrange the formula to \( I = \frac{V}{R} \) for calculating current.

In our example with a 1.5-V dry cell, we focus on ensuring it can deliver at least 25 A of current. This mathematical approach allows us to explore the capabilities and limitations of the cell while understanding how internal resistance impacts current flow.

Voltage

Voltage, also known as electric potential difference, measures the energy difference between two points in a circuit. It's described in volts (V) and is the driving force that allows current to flow. Voltage can be thought of as the "push" that makes the current move in the circuit.

Key points about voltage:

• It tells us how much energy per charge is available to push electrons through the circuit.
• Higher voltage implies a stronger "push" to move electrons through a conductor.
• It is critical when determining how much power a device can supply to a load.

In the scenario of the 1.5-V dry cell, the 1.5 volts indicate the potential energy available to move charges through the ammeter. Understanding the battery voltage is vital in deciding whether it meets the demand of providing at least 25 A. Such calculations enable us to efficiently match devices with their power sources or troubleshoot when performance issues arise.