Problem 81 - University Physics with Modern Physics

Question

A \(12.0-\mathrm{V}\) battery has an internal resistance of 0.24 \(\mathrm{s}\) and a capacity of 50.0 \(\mathrm{A} \cdot \mathrm{h}\) (see Exercise 25.47\() .\) The battery is charged by passing a 10 -A current through it for 5.0 \(\mathrm{h}\) . (a) What is the terminal voltage during charging? (b) What total electricalenergy is supplied to the battery during charging? (c) What electrical energy is dissipated in the internal resistance during charging? (d) The battery is now completely discharged through a resistor, again with a constant current of 10 \(\mathrm{A}\) . What is the external circuit resistance? (e) What total electrical energy is supplied to the external resistor? (f) What total electrical energy is dissipated in the internal resistance? (g) Why are the answers to parts (b) and (e) not the same?

Step-by-Step Solution

Answer

The terminal voltage during charging is 14.4 V, total energy supplied is 2.592 x 10^6 J, and energy dissipated in internal resistance is 4.32 x 10^5 J. External circuit resistance is 0.96 Ω, external energy supplied is 1.728 x 10^6 J, and energy dissipated in internal resistance during discharge is 4.32 x 10^5 J.

Step 1: Calculate Terminal Voltage During Charging

To find the terminal voltage \( V_t \) during charging, use the formula \( V_t = E + Ir \), where \( E = 12.0 \ \text{V} \) is the emf of the battery, \( I = 10 \ \text{A} \) is the charging current, and \( r = 0.24 \ \Omega \) is the internal resistance. Substituting the values gives: \[ V_t = 12.0 + (10)(0.24) = 14.4 \ \text{V} \] So, the terminal voltage during charging is \( 14.4 \ \text{V} \).

Step 2: Calculate Total Electrical Energy Supplied

The total electrical energy \( W_{supplied} \) supplied to the battery is given by \( V_t \cdot I \cdot t \), where \( t = 5.0 \ \text{h} \). Convert hours to seconds: \( t = 5.0 \times 3600 \ \text{s} = 18000 \ \text{s} \).Using \( V_t = 14.4 \ \text{V} \), \( I = 10 \ \text{A} \):\[ W_{supplied} = 14.4 \times 10 \times 18000 = 2.592 \times 10^6 \ \text{J} \]The total electrical energy supplied is \( 2.592 \times 10^6 \ \text{J} \).

Step 3: Calculate Energy Dissipated in Internal Resistance

The energy \( W_{dissipated} \) dissipated in the internal resistance is given by \( I^2 \times r \times t \).Substitute \( I = 10 \), \( r = 0.24 \), and \( t = 18000 \ \text{s} \):\[ W_{dissipated} = 10^2 \times 0.24 \times 18000 = 4.32 \times 10^5 \ \text{J} \]The energy dissipated in the internal resistance is \( 4.32 \times 10^5 \ \text{J} \).

Step 4: Determine External Circuit Resistance

When the battery is completely discharged with \( I = 10 \ \text{A} \), the terminal voltage is reduced due to the internal resistance, defined by \( V_t = E - Ir \). To find the external resistance \( R \), use Ohm’s law \( V = IR \) where \( V = V_t \):\[ R = \frac{V_t}{I} = \frac{12.0 - (10 \times 0.24)}{10} = \frac{9.6}{10} = 0.96 \ \Omega \]The external circuit resistance is \( 0.96 \ \Omega \).

Step 5: Calculate Energy Supplied to External Resistor

To find the total electrical energy \( W_{external} \) supplied to the external resistor, use \( I^2R \cdot t \). With \( R = 0.96 \ \Omega \), \( I = 10 \ \text{A} \), and \( t = 18000 \ \text{s} \):\[ W_{external} = 10^2 \times 0.96 \times 18000 = 1.728 \times 10^6 \ \text{J} \]The electrical energy supplied to the external resistor is \( 1.728 \times 10^6 \ \text{J} \).

Step 6: Calculate Energy Dissipated in Internal Resistance During Discharge

The energy \( W_{dissipated\_discharge} \) in the internal resistance during discharge is still \( I^2 \times r \times t \):\[ W_{dissipated\_discharge} = 10^2 \times 0.24 \times 18000 = 4.32 \times 10^5 \ \text{J} \]This energy remains the same as it depends on the current and internal resistance alone.

Step 7: Understand Difference in Supplied Energy

The difference in total electrical energy between parts (b) and (e) arises because some energy is dissipated as internal heat within the battery during the charging and discharging processes due to its internal resistance, which affects the amount of energy delivered to the external circuit.

Key Concepts

Internal Resistance

Every battery has something called internal resistance. It's a tiny resistance within the battery itself that can affect how well it works. Just like turning your bike pedal can feel harder if the chain isn't oiled, this internal resistance can cause some energy to be lost inside the battery.
This resistance can change how much voltage is actually available to your devices. It's measured in ohms (\(\Omega\)) and in this exercise, the battery's internal resistance is given as \(0.24 \ \Omega\).

• When current flows through the battery, some energy is lost due to this internal resistance.
• This lost energy appears as heat, warming up the battery slightly.
• Internal resistance causes a voltage drop inside the battery.

This means the battery's job isn't just about sending out power; it also involves managing these small losses along the way.

Ohm's Law

Ohm's Law is like the alphabet of electrical circuits. It helps us understand the relationship between voltage, current, and resistance in a simple equation: \(V = IR\).
Here, \(V\) represents voltage, \(I\) stands for current (measured in amps), and \(R\) is the resistance (ohms).

• When you know two values among current, voltage, and resistance, you can always find the third one using this formula.
• It's a fundamental tool for analyzing electric circuits.
• This law shows us how increasing the resistance or current can affect the voltage across a component.

In practical terms, if the battery's resistance increases, it can lead to a decrease in the voltage available for any gadget you're powering. Using Ohm's Law, you can calculate important values such as current flowing through a circuit or voltage needed to maintain that flow.

Electrical Energy

Electrical energy is the lifeblood of electric circuits, just like fuel is for cars. It's what's created or used up when electric current flows through a circuit.
This energy can be used in various forms, such as lighting a bulb or running a motor. In this specific case, we calculate the electrical energy involved during charging and discharging the battery.

• Energy is measured in Joules (J).
• When charging the battery, energy is supplied to it, calculated as \(W = V_t \times I \times t\).
• During discharge, energy is consumed by connected devices like resistors.

It's important to note that not all energy ends up doing useful work. Some can be lost as heat due to the internal resistance of the battery. This leads us to the understanding that the energy given by the battery and the one actually used by devices can differ.

Terminal Voltage

Terminal voltage is the voltage "available" at the battery's terminals when a device is connected and current flows through it.
It's like the actual horsepower you feel on the wheels of a car, not what's claimed by the engine.
When a battery is under load (i.e., when it's powering a circuit), the terminal voltage isn't just the battery's voltage rating \(E\) but is influenced by its internal resistance \(r\).

• The formula to calculate terminal voltage is \(V_t = E - Ir\).
• This formula shows how internal resistance can reduce the effective voltage.
• During charging, the terminal voltage is slightly different due to the direction of current flow, given by \(V_t = E + Ir\).

Understanding terminal voltage is crucial in ensuring that the devices powered by batteries are receiving enough voltage to function properly, without too much energy loss.