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Current Electricity - Potentiometer and Cells

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Internal Resistance of a Cell: Every cell has an inherent resistance to the flow of current within it, denoted as rr. This results in a terminal voltage VV that is less than the EMF EE when current II is being drawn, expressed as V=E−IrV = E - Ir.

Circuit symbol for a cell showing EMF E and internal resistance r in series.
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Principle of a Potentiometer: It works on the principle that for a uniform wire carrying a constant current, the potential drop across any segment of the wire is directly proportional to its length, i.e., V∝lV \propto l or V=klV = kl, where kk is the potential gradient.

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Cells in Series: When cells are connected in series, the equivalent EMF is the sum of individual EMFs, and the equivalent internal resistance is the sum of individual internal resistances.

Two cells connected in series.
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Cells in Parallel: When identical cells are connected in parallel, the equivalent EMF remains the same as a single cell, but the equivalent internal resistance decreases, following the reciprocal rule for parallel resistors.

Two identical cells connected in parallel.
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Sensitivity of Potentiometer: The sensitivity of a potentiometer can be increased by decreasing the potential gradient kk. This is achieved by either increasing the length of the wire or decreasing the current in the primary circuit.

Representation of a long potentiometer wire to show increased sensitivity.

📐Formulae

V=E−IrV = E - Ir

I=ER+rI = \frac{E}{R + r}

Eeq=E1+E2+...+En (Series connection)E_{eq} = E_1 + E_2 + ... + E_n \text{ (Series connection)}

req=r1+r2+...+rn (Series connection)r_{eq} = r_1 + r_2 + ... + r_n \text{ (Series connection)}

Eeq=∑Eiri∑1ri (Parallel connection)E_{eq} = \frac{\sum \frac{E_i}{r_i}}{\sum \frac{1}{r_i}} \text{ (Parallel connection)}

V=kl where k=potential gradientV = kl \text{ where } k = \text{potential gradient}

E1E2=l1l2\frac{E_1}{E_2} = \frac{l_1}{l_2}

r=R(l1l2−1)r = R \left( \frac{l_1}{l_2} - 1 \right)

💡Examples

Problem 1:

A cell of EMF EE is balanced against a length of 350 cm350 \text{ cm} on a potentiometer wire. When a resistance of 10Ω10 \Omega is connected across the cell, the balancing length becomes 300 cm300 \text{ cm}. Calculate the internal resistance of the cell.

Solution:

Given l1=350 cml_1 = 350 \text{ cm}, l2=300 cml_2 = 300 \text{ cm}, and R=10ΩR = 10 \Omega. Using the formula r=R(l1l2−1)r = R \left( \frac{l_1}{l_2} - 1 \right), we get r=10(350300−1)=10(76−1)=10×16=53Ω≈1.67Ωr = 10 \left( \frac{350}{300} - 1 \right) = 10 \left( \frac{7}{6} - 1 \right) = 10 \times \frac{1}{6} = \frac{5}{3} \Omega \approx 1.67 \Omega.

Explanation:

The internal resistance is found by comparing the balancing length of the cell in an open circuit (l1l_1) to the balancing length when shunted by a known resistor (l2l_2).

Problem 2:

In a potentiometer arrangement, a cell of EMF 1.25 V1.25 \text{ V} gives a balance point at 35.0 cm35.0 \text{ cm} length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm63.0 \text{ cm}, what is the EMF of the second cell?

Solution:

Using the principle E1E2=l1l2\frac{E_1}{E_2} = \frac{l_1}{l_2}, we have E1=1.25 VE_1 = 1.25 \text{ V}, l1=35.0 cml_1 = 35.0 \text{ cm}, and l2=63.0 cml_2 = 63.0 \text{ cm}. Therefore, E2=E1×l2l1=1.25×63.035.0=1.25×1.8=2.25 VE_2 = E_1 \times \frac{l_2}{l_1} = 1.25 \times \frac{63.0}{35.0} = 1.25 \times 1.8 = 2.25 \text{ V}.

Explanation:

Since the potential gradient kk is constant for the same potentiometer setup, the ratio of the EMFs is equal to the ratio of their respective balancing lengths.

Problem 3:

Two cells of EMFs E1E_1 and E2E_2 are connected in series to assist each other and then in series to oppose each other. The balance points are found at 500 cm500 \text{ cm} and 100 cm100 \text{ cm} respectively. Calculate the ratio E1E2\frac{E_1}{E_2}.

Schematic showing two cells in assisting (same polarity) and opposing (opposite polarity) configurations.

Solution:

In the 'assist' mode, the total EMF is E1+E2E_1 + E_2. In the 'oppose' mode, the total EMF is E1−E2E_1 - E_2. According to the potentiometer principle: E1+E2=k(500)E_1 + E_2 = k(500) E1−E2=k(100)E_1 - E_2 = k(100) Dividing the two equations: E1+E2E1−E2=500100=5\frac{E_1 + E_2}{E_1 - E_2} = \frac{500}{100} = 5 E1+E2=5E1−5E2E_1 + E_2 = 5E_1 - 5E_2 6E2=4E16E_2 = 4E_1 E1E2=64=1.5\frac{E_1}{E_2} = \frac{6}{4} = 1.5

Explanation:

This is the comparison of EMFs using the sum and difference method. The ratio of the EMFs is derived from the ratio of the balancing lengths for the combined configurations.