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Current Electricity - Potentiometer and Cells

Grade 12CBSEPhysics

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

🔑Concepts

Electromotive Force (EMF): The potential difference across the terminals of a cell when no current is being drawn from it (I=0I = 0). It is denoted by EE.

Terminal Potential Difference (VV): The potential difference across the terminals of a cell when current II flows through the circuit. It is related to EMF by V=EIrV = E - Ir, where rr is the internal resistance.

Internal Resistance (rr): The resistance offered by the electrolyte and electrodes inside a cell to the flow of current. It depends on the nature of the electrolyte, area of electrodes, and distance between them.

Cells in Series: When cells are connected in series, the equivalent EMF is the sum of individual EMFs (Eeq=E1+E2E_{eq} = E_1 + E_2) and equivalent internal resistance is req=r1+r2r_{eq} = r_1 + r_2.

Cells in Parallel: For two cells connected in parallel, the equivalent EMF is Eeq=E1r2+E2r1r1+r2E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} and the equivalent internal resistance is 1req=1r1+1r2\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}.

Potentiometer Principle: The potential drop across any portion of a wire of uniform cross-section is directly proportional to the length of that portion (VlV \propto l), provided a constant current flows through it. This gives the potential gradient k=VLk = \frac{V}{L}.

Comparison of EMFs: Two cells of EMFs E1E_1 and E2E_2 can be compared using a potentiometer by finding their respective balancing lengths l1l_1 and l2l_2, giving E1E2=l1l2\frac{E_1}{E_2} = \frac{l_1}{l_2}.

Measurement of Internal Resistance: A potentiometer can measure the internal resistance rr of a cell using the formula r=R(l1l21)r = R \left( \frac{l_1}{l_2} - 1 \right), where l1l_1 is the open circuit balancing length and l2l_2 is the closed circuit balancing length with external resistance RR.

Sensitivity of Potentiometer: A potentiometer is more sensitive if it has a smaller potential gradient kk. This can be achieved by increasing the length of the potentiometer wire or decreasing the current in the primary circuit.

📐Formulae

V=EIrV = 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=Eiri1ri (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(l1l21)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(l1l21)r = R \left( \frac{l_1}{l_2} - 1 \right), we get r=10(3503001)=10(761)=10×16=53Ω1.67Ωr = 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.

Potentiometer and Cells - Revision Notes & Key Formulas | CBSE Class 12 Physics