Question

In the experiment of potentiometer wire \(\mathrm{AB}\) is \(100 \mathrm{~cm}\) long shown in figure When \(\mathrm{AC}=40 \mathrm{~cm}\), no deflection occurs in the galvanometer. What is the value of \(R\) ? (A) \(15 \Omega\) (B) \(18 \Omega\) (C) \(12 \Omega\) (D) \(14 \Omega\)

Short answer

The value of R is \(15\Omega\), which corresponds to option (A).

Step-by-step solution

Understanding the potentiometer experiment

In the potentiometer experiment, a null-point (no deflection) in the galvanometer indicates that the potential difference across the known length of the wire (AC in this case) and the unknown resistance (R) is equal. Thus, we can set up an equation: \( \frac{V_{AC}}{AC} = \frac{V_{BR}}{BR} \) Where V_AC is the potential difference across the length AC, V_BR is the potential difference across length BR, and BR is the length of the wire between points B and R.

Determine the lengths of AC and BR

Since AC is given as 40 cm and the total length of the wire, AB, is 100 cm, we can find the length of BR as: \( BR = AB - AC \) \( BR = 100cm - 40cm = 60cm \)

Calculate the potential difference across AC and BR

As there is no deflection in the galvanometer, the potential difference across AC and R, V_AC, is equal to the potential difference across BR, V_BR. Therefore, we can write the initial equation as: \( \frac{V}{40} = \frac{V}{60} \) where V is the potential difference across both AC and BR. Solving for V, we get: \( V = \frac{2}{3} V \)

Find the value of R using Ohm's law

Using Ohm's law, we can write the equation for the potential difference across the unknown resistance R as: \( V_{BR} = I \times R \) where I is the current flowing through the circuit. Substituting the expression for V_BR from step 3, we get: \( \frac{2}{3}V = I \times R \) Since the potential difference across both AC and BR is equal, we can write another equation for the potential difference using Ohm's law for length AC: \( V_{AC} = I \times 10 \) Where the resistance of AC is \( 10\Omega \) (given in the problem).

Solve for R

Now, we are given that the potential difference between A and C is zero. This means that the potential difference is the same for both lengths of the wire AC and BR. Therefore, \( V_{AC}= V_{BR} \) . Hence, we can write our equation as: \( I \times 10 = \frac{2}{3}V = I \times R \) Now, divide both sides by I: \( 10 = \frac{2}{3}R \) Multiplying both sides by 3 and dividing by 2, we get: \( R = \frac{3}{2} \times 10 = 15\Omega \) Hence, the value of R is 15Ω, which corresponds to option (A).

Key concepts

Null Point in Potentiometer Experiment

In the context of a potentiometer experiment, the 'null point' is a crucial concept that signifies a specific balance in the setup. At this point, the galvanometer connected in the circuit shows no deflection. This means that the potential difference across the selected length of wire is balanced with the potential drop across the unknown resistor.

Finding the null point ensures that:

• The potential difference across the circuit segments is perfectly matched
• No current flows through the galvanometer indicating the balance is achieved
• The wire's length ratio helps determine the unknown resistance's value

When the galvanometer reads zero, it allows for the precise calculation of an unknown resistance, like in our exercise. By understanding and locating the null point, we not only achieve accuracy but optimize the potentiometer's reliability in measuring voltage differences.

Galvanometer's Role in the Experiment

A galvanometer is a sensitive instrument used to detect and measure small electric currents. In the potentiometer experiment, its role is pivotal for indicating whether the circuit is balanced. The term 'no deflection' means exactly that there is no current flowing through the galvanometer at the null point, signifying balance.

Key notes on the use of a galvanometer include:

• Detects minute currents with high sensitivity.
• Gives a zero reading when the null point is reached.
• Helps adjust the circuit for precision resistance measurement.

When adjusting the potentiometer, the aim is to achieve a zero deflection on the galvanometer. Doing so confirms that the voltage across the wire segment equals the voltage across the resistive load. This precise functionality makes the galvanometer indispensable to the potentiometer setup.

Understanding Ohm's Law in Potentiometer

Ohm's Law underpins the calculations in the potentiometer experiment. It's expressed as: \[ V = I R \]This equation correlates current (I), voltage (V), and resistance (R). For the potentiometer, Ohm's Law translates to how potential differences across the wire segments aid in determining unknown resistance.

Applying Ohm's Law involves:

• Relating the potential difference across wire segments to their lengths.
• Solving equations to find the unknown resistance based on current and known values.
• Using proportional relationships between segment lengths to simplify the calculations.

In our exercise, once the null point was determined, Ohm's Law facilitated the calculation of resistance R. By setting up the proportionate relationships and applying Ohm's Law correctly, students can solve for unknown values with accuracy, illustrating the law's practical application in experimental physics.