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  • Two cells of emf and  <img alt="(E_1 > E_2 )" src="https://learn.careers360.com/latex-image/?%28E

Two cells of emf E_1 and E_{2} (E_1 > E_2 ) are connected as shown in the figure below. When a potentiometer is used to measure potential difference between the points A and B, the balancing length of the potentiometer wire is 300 cm. But the same potentiometer for the potential difference between points A and C, gives the balancing length 100 cm. Find \frac{E_{1}}{E_{2}} .

 

 

 
 
 
 
 

Answers (1)

\\E_1\propto300\\E_1-E_2\propto100\\\\\frac{E_1-E_2}{E_1}=\frac{1}{3}\\\\1-\frac{E_2}{E_1}=\frac{1}{3}\\\\\frac{E_2}{E_1}=\frac{2}{3}\\\\\frac{E_1}{E_2}=\frac{3}{2}

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