Question

A battery of \(2 \mathrm{~V}\) and internal resistance 1 is used to send a current through a potentiometer wire of length \(200 \mathrm{~cm}\) and resistance \(4 \mathrm{Q}\) What is the potential gradient of the wire? (A) \(8 \times 10^{-3} \mathrm{~V} \mathrm{~cm}^{-1}\) (B) \(4 \times 10^{-3} \mathrm{~V} \mathrm{~cm}^{-1}\) (C) \(6 \times 10^{-3} \mathrm{~V} \mathrm{~cm}^{-1}\) (D) \(2 \times 10^{-3} \mathrm{~V} \mathrm{~cm}^{-1}\)

Short answer

The potential gradient of the wire is \(8 \times 10^{-3}~V~cm^{-1}\).

Step-by-step solution

Identify given values from the problem

The exercise provides us with the following information: - The battery has a voltage of \(2~V\) and an internal resistance of \(1~\Omega\). - The potentiometer wire has a length of \(200~cm\) (or \(2~m\)) and a resistance of \(4~\Omega\).

Calculate the total resistance of the system

The total resistance of the system is the sum of the internal resistance of the battery and the resistance of the potentiometer wire: \[R_{total} = R_{battery} + R_{wire}\] \[R_{total} = 1~\Omega + 4~\Omega\] \[R_{total} = 5~\Omega\]

Determine the current flowing through the system

Using Ohm's law, calculate the current flowing through the system (I): \[V = I \cdot R\] \[I = \frac{V}{R_{total}}\] Substitute the values of V and \(R_{total}\): \[I = \frac{2~V}{5~\Omega}\] \[I = 0.4~A\]

Calculate the potential drop across the potentiometer wire

We will use Ohm's law to determine the potential drop (\(V_{wire}\)) across the potentiometer wire length: \[V_{wire} = I \cdot R_{wire}\] Substitute the values of I and \(R_{wire}\): \[V_{wire} = 0.4~A \cdot 4~\Omega\] \[V_{wire} = 1.6~V\]

Calculate the potential gradient of the wire

The potential gradient (\(k\)) is the potential drop per unit length of the wire: \[k = \frac{V_{wire}}{length}\] Substitute the values of \(V_{wire}\) and length: \[k = \frac{1.6~V}{200~cm}\] \[k = 8 \times 10^{-3}~V~cm^{-1}\] The potential gradient of the wire is \(8 \times 10^{-3}~V~cm^{-1}\), which corresponds to answer (A).

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in physics and electrical engineering.It describes the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit.According to Ohm's Law, the voltage across a resistor is directly proportional to the current flowing through it and the resistance of the resistor itself.

The mathematical representation of Ohm's Law is:

• \( V = I \times R \)

This equation means that by knowing any two of the three variables (voltage, current, resistance), we can calculate the third.In our exercise, we used Ohm's Law to find the current flowing through the circuit.Understanding this law is crucial for solving many electricity problems.It helps us to determine how a circuit will respond to different input conditions.Ohm's Law is the starting point for learning more complex circuit analysis techniques.

Internal Resistance

Internal resistance is an intrinsic property of all power sources such as batteries.It represents the opposition to the flow of current within the battery itself.

This resistance affects the efficiency of power delivery to an external circuit because some of the energy is lost as heat inside the battery.For an ideal battery, internal resistance would be zero, meaning no energy loss.However, in real-world applications, internal resistance is always present and needs to be considered in calculations.

• The internal resistance can cause the terminal voltage of a battery to drop under load.
• It can be modeled as a resistor within the battery's circuit in series with the external circuit.

In the exercise, the internal resistance of 1 \( \, \Omega \) was added to the potentiometer wire's resistance to find the total resistance.This total resistance was then used to compute the current through the circuit using Ohm's Law.

Potentiometer Wire

A potentiometer wire is a type of resistor that can help measure potential difference across a circuit.It is often used in experiments due to its ability to provide a known resistance over a certain length.

• The resistance of a potentiometer wire is related to its material, length, and cross-sectional area.
• It allows for the precise measurement of battery EMF without drawing current from it, making it ideal for delicate measurements.

In this exercise, a potentiometer wire with a length of 200 cm and resistance of 4 \( \, \Omega \) was used.We calculated the potential gradient along the wire, which tells us how much the voltage changes per unit length.This is a vital step to understanding how voltage distributes along different segments of the wire, making the potentiometer wire a powerful tool in circuit analysis.

Current Calculation

Current calculation is an essential step in understanding how electricity flows through a circuit.In this exercise, we needed to calculate the current to determine the potential drop across the potentiometer wire.

• By using the expression \( I = \frac{V}{R} \), where \( V \) is voltage and \( R \) is resistance, we calculated the current flowing through the total resistance of the circuit.
• This calculated current then assisted in finding the potential difference across specific sections of the circuit.

Determining the current is often a precursor to further analysis, like finding voltage drops and power consumption in various parts of the circuit.In the given exercise, the current flowing through the system was determined to be \( 0.4 \) amperes, which was pivotal for subsequent calculations.