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Question

A potentiometer wire of length L and a resistance r are connected in series with a battery of emf E0 and a resistance r1. An unknown emf E is balanced at a length l of the potentiometer wire. The emf E will be:

The correct answer is

(E0 r) / (r + r1) * (l / L)

Understanding the Potentiometer Principle

A potentiometer is a versatile electrical instrument primarily used for measuring the electromotive force (emf) of a cell or for comparing emfs. Its working is based on the principle that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any portion of the wire is directly proportional to its length.

Primary Circuit Analysis of the Potentiometer

In the given setup, the primary circuit of the potentiometer consists of the battery with emf \(E_0\), the potentiometer wire of length \(L\) and resistance \(r\), and an external series resistance \(r_1\). These components are connected in series. The total resistance of the primary circuit is the sum of the resistance of the potentiometer wire and the external series resistance. Total resistance, \(R_{total} = r + r_1\). The current flowing through the primary circuit can be calculated using Ohm's Law, considering the total emf and the total resistance in the circuit. Current in the primary circuit, \(I = \frac{\text{Total emf}}{\text{Total resistance}}\) \(I = \frac{E_0}{r + r_1}\)

Potential Drop Across the Potentiometer Wire

The potential drop across the entire length \(L\) of the potentiometer wire is the voltage across its resistance \(r\) when the current \(I\) flows through it. Potential drop across the wire, \(V_{wire} = I \times r\) Substituting the expression for \(I\): \(V_{wire} = \left(\frac{E_0}{r + r_1}\right) \times r = \frac{E_0 r}{r + r_1}\)

Calculating the Potential Gradient

The potential gradient (k) is the potential drop per unit length of the potentiometer wire. It is calculated by dividing the potential drop across the entire wire by its total length. Potential gradient, \(k = \frac{V_{wire}}{L}\) Substituting the expression for \(V_{wire}\): \(k = \frac{\frac{E_0 r}{r + r_1}}{L} = \frac{E_0 r}{L(r + r_1)}\) The potential gradient \(k\) has units of Volts per meter (V/m) or Volts per centimeter (V/cm), depending on the unit of length used.

Balancing the Unknown Emf E

An unknown emf \(E\) is connected in the secondary circuit of the potentiometer in opposition to the primary circuit's potential drop. A galvanometer is used to detect the flow of current. The balancing length \(l\) is the length of the potentiometer wire from the positive terminal end where the galvanometer shows zero deflection. This zero deflection condition means no current is flowing through the secondary circuit, indicating that the potential drop across the length \(l\) of the potentiometer wire is exactly equal to the unknown emf \(E\). Potential drop across length \(l\) = Potential gradient \(\times\) length \(l\) Potential drop across length \(l\) = \(k \times l\) Since the unknown emf \(E\) is balanced by the potential drop across the length \(l\): \(E = k \times l\) Substituting the expression for the potential gradient \(k\): \(E = \left(\frac{E_0 r}{L(r + r_1)}\right) \times l\) This can be rearranged as: \(E = \frac{E_0 r}{r + r_1} \times \frac{l}{L}\) This equation gives the value of the unknown emf \(E\) in terms of the known parameters of the primary circuit (\(E_0\), \(r\), \(r_1\), \(L\)) and the measured balancing length \(l\).

Final Expression for Unknown Emf

Based on our derivation, the unknown emf \(E\) is given by the formula: \(E = \frac{E_0 r}{r + r_1} \times \frac{l}{L}\) Let's compare this with the given options. The derived expression matches one of the provided options.

Revision Table: Key Concepts

Concept Description Formula/Relationship
Potentiometer Principle Potential drop across a uniform wire is proportional to length when current is constant. \(V \propto l\)
Primary Circuit Current (I) Current through the potentiometer wire and series resistance. \(I = \frac{E_0}{r + r_1}\)
Potential Drop across Wire (\(V_{wire}\)) Total voltage across the length L of the potentiometer wire. \(V_{wire} = I \times r = \frac{E_0 r}{r + r_1}\)
Potential Gradient (k) Potential drop per unit length of the wire. \(k = \frac{V_{wire}}{L} = \frac{E_0 r}{L(r + r_1)}\)
Balancing Condition Unknown emf equals potential drop across balancing length. \(E = k \times l\)

Additional Information on Potentiometers

Potentiometers are preferred over voltmeters for measuring emf because they draw no current from the cell at the point of balance, thus measuring the true emf rather than terminal voltage. The sensitivity of a potentiometer can be increased by decreasing the potential gradient along the wire. This can be achieved by increasing the length of the potentiometer wire or by increasing the resistance \(r_1\) in the primary circuit, which reduces the current and thus the potential drop across the wire. Potentiometers can also be used to compare resistances and internal resistances of cells.
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Important Questions from Dual Nature of Radiation and Matter

  1. The work function for an Aluminium surface is 4.2 eV. Find the threshold wavelength for the photoelectric emission.

  2. The time taken by light to travel normally through a glass plate of thickness 1 mm would be:

    (Take refractive index of glass = 1.5)

  3. Energy of a photon corresponding to a wavelength of 600 nm is 2.08 eV. The energy of a photon of wavelength 400 nm will be:

  4. A particle moves three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10-4. The mass of the particle is:

  5. According to Einstein’s photoelectric equation, the plot of the Kinetic Energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope:

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