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  • Two cells of emfs <img alt="\mathrm{\varepsilon_1 \: and \: \: \varepsilon_2\left(\varepsilon_1>\varepsilon_2\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cvarepsi

Two cells of emfs \mathrm{\varepsilon_1 \: and \: \: \varepsilon_2\left(\varepsilon_1>\varepsilon_2\right)} are connected as shown in figure. When a potentiometer is connected between \mathrm{A}$ and $\mathrm{B}, the balancing length of the potentiometer wire is 300 \mathrm{~cm}. On connecting the same potentiometer between \mathrm{\mathrm{A} \: and \: \mathrm{C}}, the balancing length is 100 \mathrm{cm}. The ratio \frac{\varepsilon_1}{\varepsilon_2} is:

Option: 1

3:1


Option: 2

1:3


Option: 3

2:3


Option: 4

3:2


Answers (1)

best_answer

When potentiometer is connected between \mathrm{A \: and \: B}, then it measures only \varepsilon _1 and when connected between \mathrm{A\: and \: \mathrm{C}}, then it measures \varepsilon_1-\varepsilon_2.\varepsilon_1-\varepsilon_2.\varepsilon _1-\epsilon _2.

\therefore \quad \frac{\varepsilon_1}{\varepsilon_1-\varepsilon_2}=\frac{l_1}{l_2}, \quad \frac{\varepsilon_1-\varepsilon_2}{\varepsilon_1}=\frac{l_2}{l_1}

\text { or } \quad 1-\frac{\varepsilon_2}{\varepsilon_1}=\frac{100}{300} \quad \text { or } \quad \frac{\varepsilon_2}{\varepsilon_1}=1-\frac{1}{3} \quad \text { or } \quad \frac{\varepsilon_2}{\varepsilon_1}=\frac{2}{3} \quad \text { or } \quad \frac{\varepsilon_1}{\varepsilon_2}=\frac{3}{2}

Hence option 4 is correct.

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