Problem 78 - Physics for Scientists and Engineers with Modern Physics

Question

A battery produces \(40.8 \mathrm{~V}\) when \(7.40 \mathrm{~A}\) is drawn from it, and \(47.3 \mathrm{~V}\) when \(2.80 \mathrm{~A}\) is drawn. What are the \(\mathrm{emf}\) and internal resistance of the battery?

Step-by-Step Solution

Answer

The emf is approximately 51.26 V and the internal resistance is approximately 1.413 ohms.

Step 1: Understand the Problem

We are given two separate cases of voltage output from a battery with different current draws. We have: \( V_1 = 40.8 \text{ V} \) with \( I_1 = 7.40 \text{ A} \) and \( V_2 = 47.3 \text{ V} \) with \( I_2 = 2.80 \text{ A} \). We aim to find the electromotive force (emf, \( E \)) and the internal resistance (\( r \)) of the battery.

Step 2: Write the Voltage Formula

The battery's terminal voltage \( V \) is given by the formula: \( V = E - I \cdot r \), where \( E \) is the emf and \( r \) is the internal resistance.

Step 3: Set Up Equations

Using the given data, we can set up two equations based on the formula: \( 40.8 = E - 7.40 \cdot r \) for the first situation, and \( 47.3 = E - 2.80 \cdot r \) for the second.

Step 4: Solve the System of Equations

We solve the equations simultaneously to find \( E \) and \( r \). Subtract the first equation from the second: \( (47.3 - 40.8) = (E - 2.80 \cdot r) - (E - 7.40 \cdot r) \), leading to \( 6.5 = 4.60 \cdot r \). Thus, \( r = \frac{6.5}{4.60} \).

Step 5: Calculate Internal Resistance

Calculate \( r \): \( r = \frac{6.5}{4.60} ≈ 1.413 \text{ ohms} \).

Step 6: Calculate EMF

Substitute \( r \) back into one of the original equations to find \( E \). For example, use \( 40.8 = E - 7.40 \cdot 1.413 \): \[ E = 40.8 + 7.40 \times 1.413 \].

Step 7: Solve for EMF

Solve for \( E \): \( E = 40.8 + 10.4562 = 51.2562 \text{ V} \).

Step 8: Verify Solution

Finally, verify that both values satisfy both original equations: \( 40.8 \approx 51.2562 - 7.40 \times 1.413 \), and \( 47.3 \approx 51.2562 - 2.80 \times 1.413 \). Both hold true, confirming correctness.

Key Concepts

Internal Resistance

Internal resistance is the resistance within a battery or cell that causes a drop in the voltage output when a current is flowing. It's a crucial factor because it affects the efficiency of the battery. The more internal resistance a battery has, the more the voltage will drop under load, leading to less power being delivered to the connected circuit.
To quantify internal resistance, we use the formula derived from Ohm's Law:

• Terminal Voltage (V) = EMF (E) - Current (I) \( \times \) Internal Resistance (r)

In this formula, when current flows, the internal resistance \( r \) causes a voltage drop \( I \times r \). By solving for \( r \) using given measurements of current and voltage, we can determine how the battery performs under different loads.
This resistance is a product of the physical and chemical properties of the materials in the battery and is generally considered an undesired property because it "wastes" some of the battery's energy as heat.

Ohm's Law

Ohm's Law is a fundamental principle that describes how voltage, current, and resistance relate to each other in an electrical circuit. The law is usually expressed with the formula:

• Voltage (V) = Current (I) \( \times \) Resistance (R)

This simple formula allows us to understand and predict how electrical circuits respond to changes in voltage or resistance. The beauty of Ohm's Law is in its simplicity and applicability to almost all circuits. In the context of a battery, this principle helps calculate how changes in internal resistance affect the terminal voltage, giving us valuable information about battery condition and performance.
Furthermore, Ohm's Law is essential for circuits with multiple resistors and components, as it helps us understand the distribution of current through each part of the circuit.
For accurate predictions, it's important to consider all forms of resistance, including external resistors and internal resistive properties of the components themselves.

Simultaneous Equations

Simultaneous equations are a set of equations with multiple variables. In our case, these equations stem from measuring the voltage and current flowing from the battery at different times. The essence of using simultaneous equations is to solve two or more equations to find the unknown values—here, the electromotive force (EMF) and internal resistance of the battery.
When two conditions or measurements for a system are given, as with different current draws and their respective voltages from the battery, simultaneous equations enable the calculation of two unknowns.

• For example, the equations \(( V_1 = E - I_1 \times r )\) and \(( V_2 = E - I_2 \times r )\) were solved together to determine both E and r .

By subtracting one equation from another, we can eliminate one of the variables, allowing us to solve for the remaining unknown. This method is a powerful tool in physics and mathematics for finding solutions in systems with multiple equations and variables.