Problem 112 - Physics

Question

A \(75.0-\Omega\) and a \(45.0-\Omega\) resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is \(0.294 \mathrm{~A}\). When the \(45.0-\Omega\) resistor is disconnected, the current from the battery drops to \(0.116 \mathrm{~A}\). Determine (a) the emf and (b) the internal resistance of the battery.

Step-by-Step Solution

Answer

a) Emf is 8.98 V; b) Internal resistance is 2.42 Ω.

Step 1: Understanding Resistance in Parallel

In a parallel circuit, the total or equivalent resistance \(R_{eq}\) is given by \(\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}\). Here, \(R_{1} = 75.0 \, \Omega\) and \(R_{2} = 45.0 \, \Omega\).

Step 2: Calculate Total Resistance in Parallel

Calculate the equivalent resistance:\[ \frac{1}{R_{eq}} = \frac{1}{75.0} + \frac{1}{45.0} \]\[ \frac{1}{R_{eq}} = \frac{45 + 75}{3375} \]\[ \frac{1}{R_{eq}} = \frac{120}{3375} \]\[ R_{eq} = \frac{3375}{120} = 28.125 \, \Omega \]

Step 3: Use Ohm's Law for the Entire Circuit

When both resistors are connected, the total current is 0.294 A.By Ohm's Law \(V_{emf} = I \times R_{total}\), where \(R_{total} = R_{eq} + r\), \(r\) being the internal resistance of the battery.

Step 4: Calculate Current with One Resistor

When only the \(75.0 \, \Omega\) resistor is connected, the total resistance is \(R_{1} + r\).The current is 0.116 A.So, \(V_{emf} = 0.116 \, A \times (75.0 \, \Omega + r)\).

Step 5: Set Up Equations

Using both current situations:1. \(0.294 = \frac{V_{emf}}{28.125 + r} \)2. \(0.116 = \frac{V_{emf}}{75.0 + r} \)

Step 6: Solve the Equations

Solve these two equations simultaneously.From Equation 1:\[ V_{emf} = 0.294(28.125 + r) \]From Equation 2:\[ V_{emf} = 0.116(75.0 + r) \]Equate both expressions for \(V_{emf}\):\[ 0.294(28.125 + r) = 0.116(75.0 + r) \]Expand and simplify:\[ 8.27 + 0.294r = 8.7 + 0.116r \]\[ 0.294r - 0.116r = 8.7 - 8.27 \]\[ 0.178r = 0.43 \]\[ r = \frac{0.43}{0.178} = 2.42 \, \Omega \]

Step 7: Calculate the Emf

Substitute \(r = 2.42 \, \Omega\) back into either equation to find \(V_{emf}\).Using Equation 1:\[ V_{emf} = 0.294(28.125 + 2.42) \]\[ V_{emf} = 0.294 \times 30.545 = 8.98 \, V \]

Step 8: Conclusions

(a) The emf of the battery is approximately \(8.98 \, V\).(b) The internal resistance is approximately \(2.42 \, \Omega\).

Key Concepts

Resistance in Parallel

When we talk about resistors connected in parallel, we are dealing with a situation where multiple resistors are connected side by side in an electrical circuit. The unique feature of parallel circuits is that they provide multiple pathways for the current to travel. This configuration affects how total resistance is calculated. The total or equivalent resistance in a parallel circuit can be found using the reciprocal rule:

• Formula: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots \).
• For two resistors, such as a \(75.0\, \Omega\) and \(45.0\, \Omega\), the total equivalent resistance, \( R_{eq} \), is calculated as in the solution above.

By using this formula, students can calculate the resistance experienced in the circuit, which will be less than the smallest resistor in parallel. This unique behavior of parallel circuits is due to multiple paths allowing for easier flow of current.

Equivalent Resistance

The concept of equivalent resistance is crucial for simplifying complex circuits. When multiple resistors are combined in a circuit, they can often be replaced by a single resistor that has the same effect on the circuit's overall resistance. For parallel circuits, the equivalent resistance \( R_{eq} \) is always lower than the smallest individual resistance in the configuration. This contrasts with series circuits, where resistances add up.
Understanding equivalent resistance is important because it allows us to analyze and design circuits more efficiently. Knowing how to calculate \( R_{eq} \) through the reciprocal method helps in predicting how a circuit will behave when different resistors are rearranged, added, or removed.

Internal Resistance

Internal resistance is a property of batteries and other power sources that opposes the flow of current within the device itself. Every battery or cell comes with an inherent resistance, which slightly reduces the voltage available for the external circuit. This internal resistance \( r \) can affect both the current and terminal voltage when the battery is connected to a load. In the exercise, we determined this by comparing two different scenarios with variations in current, allowing us to deduce this hidden resistance.
The knowledge of internal resistance is crucial for accurately measuring a battery's performance and efficiency, especially in high-demand applications. It helps in understanding how much energy is being lost to heat within the battery and can guide decisions in choosing power sources for specific tasks.

Electromotive Force (emf)

Electromotive force, commonly known as emf, is the energy provided by a battery or power source per Coulomb of charge. It is the maximum potential difference the battery can provide when no current flows, often thought of as the ideal voltage of the battery. In practical scenarios, the terminal voltage is often less than the emf because of internal resistance. However, when calculating the emf, it is crucial to consider under different load conditions, as seen in the exercise.
Understanding emf involves realizing it is not a force but rather a potential energy difference per unit charge. By using Ohm's Law and current calculations, we can solve for both the emf and its effects. This can help determine how much potential energy the battery can offer under ideal conditions.