Question

A Potentiometer wire, \(10 \mathrm{~m}\) long, has a resistance of \(40 \Omega\). It is connected in series with a resistance box and a \(2 \mathrm{v}\) storage cell If the potential gradient along the wire is \(0.1 \mathrm{~m} \mathrm{v} / \mathrm{cm}\), the resistance unplugged in the box is. (A) \(260 \Omega\) (B) \(760 \Omega\) (C) \(960 \Omega\) (D) \(1060 \Omega\)

Short answer

The resistance unplugged in the resistance box is \(760 \Omega\), which corresponds to option (B).

Step-by-step solution

1. Understand the given information and identify what needs to be found

We are given the following information: 1. Length of the potentiometer wire (L): \(10 \mathrm{~m}\) 2. Resistance of the potentiometer wire (R): \(40 \Omega\) 3. Voltage of the storage cell (V): \(2 \mathrm{~V}\) 4. Potential gradient along the wire (K): \(0.1 \mathrm{~mV/cm}\) We need to find the resistance unplugged in the resistance box (Rb).

2. Convert potential gradient to volts per meter

The potential gradient is given in millivolts per centimeter. Let's first convert it to volts per meter. 1 mV = \(10^{-3}\) V and 1 cm = \(10^{-2}\) m So, \(K = 0.1 \cdot 10^{-3} \cdot 10^{-2} = 0.1 \cdot 10^{-1} \mathrm{V/m}\)

3. Calculate the potential difference across the wire

Potential difference across the wire (Vw) can be calculated using the potential gradient (K) and the length of the wire (L): \[Vw = K \cdot L = 0.1 \cdot 10^{-1} \cdot 10 = 0.1 \mathrm{V}\]

4. Calculate the potential difference across the resistance box

Because the resistance box is connected in series with the wire, V = Vw + Vrb, where Vrb is the potential difference across the resistance box. We can solve for Vrb: \[Vrb = V - Vw = 2 - 0.1 = 1.9 \mathrm{V}\]

5. Use Ohm's Law to find the resistance in the resistance box

Now, we can use Ohm's Law to find the resistance in the resistance box. Ohm's Law states that V = I * R, where V is the potential difference, I is the current, and R is the resistance. We know the potential difference across the wire, Vw, and its resistance, R. So, we can find the current in the circuit: \[I = \frac{Vw}{R} = \frac{0.1}{40} = \frac{1}{400} \mathrm{A}\] Now, we can use the current and the potential difference across the resistance box to find the resistance unplugged in the box (Rb): \[Rb = \frac{Vrb}{I} = \frac{1.9}{\frac{1}{400}} = 760 \Omega\] So, the resistance unplugged in the resistance box is \(760 \Omega\), which corresponds to option (B).

Key concepts

Potential Gradient

A potential gradient refers to how much voltage changes along a length of a conductor wire. Imagine it like a slope, where you're seeing how steep the change in voltage is from one end to the other. On a potentiometer wire, this tells us how much potential or voltage drops per unit length. In our exercise, the potential gradient is expressed as 0.1 mV/cm.

To solve problems, we often need to convert this value into units that match other parts of the calculation, like volts per meter (V/m). This conversion helps standardize our equations and simplifies finding other values, like the total voltage drop across the wire.

Ohm's Law

Ohm's Law is a fundamental principle in electronics and electrical engineering. It defines the relationship between voltage (V), current (I), and resistance (R) in a circuit through the equation: \[ V = I \cdot R \]

In simple terms, Ohm's Law tells us that voltage is the product of the current flowing through a material and its resistance. For our exercise, once the potential gradient and length provided the voltage across the potentiometer wire, Ohm's Law helped us calculate the current by rearranging the formula to: \[ I = \frac{V}{R} \]

This step was crucial for determining the resistance within the series circuit, such as calculating the unplugged resistance in the resistance box.

Series Circuits

In a series circuit, components are connected end-to-end, forming a single path for current flow. This means the same current flows through all components, and we add up their resistances.

For the given problem, both the potentiometer wire and the resistance box were part of a series circuit with a storage cell. This setup necessitated calculating how the supplied voltage divides across the individual components. Each part of the circuit (the wire and the resistance box) shared the total voltage provided by the storage cell.

This shared path for current flow is pivotal in solving for unknown resistances, as it allows us to apply Ohm's Law and other relations consistently across the series path.

Resistance Calculation

Calculating resistance in the context of circuits often involves deducing unknown values from the given data. In this problem, the resistance of the wire was known, and by knowing the entire setup was in series, we could use the rules of series circuits.

Using the voltage drop across known parts of a circuit allows one to back-calculate the unknown resistance. The relation based on Ohm's Law: \[ R = \frac{V}{I} \]

helped solve for the resistance in the box, once the current through the circuit was established. Knowing how to juggle these relationships is essential for effectively working through electrical circuit problems, where not all values are directly given. This knowledge helps piece together the puzzle by understanding how an electric field interacts with resistance, ultimately affecting current flow, and potential distribution.