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Expert-verified Found in: Page 777 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Calculate the ${{\mathbf{emf}}}_{{\mathbf{x}}}$of a dry cell for which a potentiometer is balanced when ${{\mathbf{R}}}_{{\mathbf{x}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{200}}{\mathbf{\Omega }}$, while an alkaline standard cell with an emf ${\mathbf{1}}{\mathbf{.}}{\mathbf{600}}{\mathbf{V}}$of requires to ${{\mathbf{R}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{247}}{\mathbf{\Omega }}$balance the potentiometer.

The of a dry cell being for which a potentiometer is balanced is ${\mathrm{E}}_{\mathrm{x}}=1.540\mathrm{V}$.

See the step by step solution

## Step 1: Concept Introduction

The electric potential generated by an electrochemical cell or a changing magnetic field is known as electromotive force. Electromotive force (EMF) is a well-known abbreviation.

## Step 2: Information Provided

• Emf of a standard cell: ${\mathrm{E}}_{\mathrm{s}}=1.600\mathrm{V}$
• Resistance of the potentiometer: ${\mathrm{R}}_{\mathrm{x}}=1.200\mathrm{\Omega }$
• Standard resistance value: ${\mathrm{R}}_{\mathrm{s}}=1.247\mathrm{\Omega }$

## Step 3: Calculation for EMF of the cell

Calculate of a cell using a potentiometer. The formula is -

${\mathrm{E}}_{\mathrm{x}}={\mathrm{E}}_{\mathrm{s}}\frac{{\mathrm{R}}_{\mathrm{x}}}{{\mathrm{R}}_{\mathrm{s}}}\phantom{\rule{0ex}{0ex}}=\left(1.600\mathrm{V}\right)\frac{1.200\mathrm{\Omega }}{1.247\mathrm{\Omega }}\phantom{\rule{0ex}{0ex}}=1.600\mathrm{V}×0.96\phantom{\rule{0ex}{0ex}}=1.540\mathrm{V}$

Therefore, the value for emf is ${\mathrm{E}}_{\mathrm{x}}=1.540\mathrm{V}$. ### Want to see more solutions like these? 