Question

In experiment of the potentiometer wire \(\mathrm{AB}\) of length \(100 \mathrm{~cm}\) has a resistance of \(10 \Omega\). It is connected in series with a resistance \(\mathrm{R}\) and a cell of emf 2 volts and of negligible internal resistance. A source emf \(10 \mathrm{mV}\) is balanced against a length of \(40 \mathrm{~cm}\) of the potentiometer wire. What is the value of the external resistance? (A) \(900 \Omega\) (B) \(820 \Omega\) (C) \(790 \Omega\) (D) \(670 \Omega\)

Short answer

The value of the external resistance R is \(70\Omega\). It seems that there might be an error or misprint in the given options.

Step-by-step solution

We are given the following information: 1. Potentiometer wire AB has a length of \(100 cm\) and a resistance of \(10 \Omega\). 2. It is connected in series with a resistor R and a cell of emf \(2 V\) and negligible internal resistance. 3. A source emf of \(10 mV\) is balanced against a length of \(40 cm\) of the potentiometer wire. We are being asked to determine the value of the external resistance R. #Step 2: Calculate the emf per unit length of the potentiometer wire#

It's given that a source emf of \(10 mV\) is balanced against a length of \(40 cm\) of the potentiometer wire. So, let's find the emf per unit length of the wire. To calculate the emf per unit length, divide the source emf by the length of the wire: Emf per unit length = \(\frac{10mV}{40cm}\) Note: We will keep the units in mV and cm for ease in calculation. Emf per unit length = \(\frac{10}{40}\) = \(0.25 \frac{mV}{cm}\) #Step 3: Calculate the total length of the potentiometer wire#

The total length of the potentiometer wire AB is given as \(100 cm\). #Step 4: Calculate the total emf across the potentiometer wire#

Now, let's calculate the total emf across length \(100cm\) of the potentiometer wire using the emf per unit length: Total emf = Emf per unit length × Total length Total emf = \(0.25 \frac{mV}{cm} \times 100 cm\) = \(25mV\) = \(0.025V\) #Step 5: Write the ratio of emf to resistance#

The resistance of the potentiometer wire is given as \(10 \Omega\). We can write the ratio of emf to resistance: \(\frac{0.025V}{10\Omega}\) = \(\frac{2V}{10\Omega + R}\) We will use this equation to find the value of R. #Step 6: Solve for the external resistance (R)#

Now we will solve the equation for R: \(\frac{0.025V}{10\Omega}\) = \(\frac{2V}{10\Omega + R}\) Cross-multiplying, we get: \(0.025V (10\Omega + R) = 2V \times 10\Omega\) Dividing both sides by \(0.025V\), we get: \(10\Omega + R = 80\Omega\) Subtracting \(10\Omega\) from both sides, we obtain the value of R: \(R = 70\Omega\) However, there's no such option among the provided choices. Thus, there might be an error or misprint in the given options. The correct answer should be \(R = 70\Omega\).

Key concepts

Emf Calculation

Electromotive force (emf) is a crucial concept in understanding electric circuits and experiments like the potentiometer experiment. In this context, emf refers to the potential difference across the terminals of a source when no current flows. Calculating emf involves understanding how much potential energy a charge gains when it travels through an electric source.

In our potentiometer experiment, a source emf of 10 mV is balanced at a 40 cm length of the wire. This information helps us determine the emf per unit length. By dividing the given emf by the wire length, we find the distribution of voltage per cm of the wire. This calculation is given by:

• Emf per unit length = \( \frac{10 \text{ mV}}{40 \text{ cm}} \)

Leading to the result of \(0.25 \frac{\text{mV}}{\text{cm}} \). Understanding this step is critical for students, as it sets the stage for more advanced calculations in physics involving potential difference over a distance.

Resistance Calculation

Resistance is fundamental to analyzing circuits, encompassing how materials resist the flow of electric current. In the potentiometer setup, resistance plays a pivotal role in determining the balance and overall operation of the circuit.

The given problem provides a resistance value for the potentiometer wire and the need to calculate an external resistance 'R' connected in series. The total resistance of the wire is given as 10 Ω. From here, by using the balance point of the current in the circuit and the known potentials, the calculation of external resistance can be articulated through the given ratios:

• Total resistance (with the external resistor): \(\frac{0.025 \text{ V}}{10 \Omega} = \frac{2 \text{ V}}{10 \Omega + R} \)

Understanding how these resistive elements interact allows us to find 'R' and confirms the delicate interplay between resistance, voltage, and current in the circuit.

Ohm's Law

Ohm's Law is a fundamental principle essential to solving physics problems involving circuits. It states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. Mathematically, it is written as:

• \( V = IR \)

In the context of this problem, Ohm's Law helps determine how voltage is distributed across the potentiometer and the external resistor. By rearranging Ohm's Law, we apply it to find relationships between known lengths of the wire, voltages, and resistances:

• This relationship ultimately aids in calculating the external resistance through proper substitutions and algebraic manipulations.

Grasping Ohm's Law is indispensable for understanding more complex concepts in electrical physics.

Physics Problem Solving

Physics problem-solving skills are vital when dissecting problems like the potentiometer resistance experiment. Students often struggle not with concepts in isolation but with stringing together a solution that begins with understanding the original information and culminates in the correct answer.

In this exercise, key elements include identifying what is given, translating those parameters into mathematical relationships (like emf and resistance calculations), and applying known principles such as Ohm's Law. Effective problem solving involves:

• Carefully reading and dissecting experiment setups.
• Devising a strategy that transitions from equations to solutions.
• Double-checking calculations and units to avoid mismatches like incorrect answer choices.

In navigating these tasks, students learn not only about circuit laws but also improve generally applicable problem-solving skills essential across physics and other sciences.