Question 595: Current Electricity

Question

Two cells having unknown emfs $E_1$ and $E_2$ $\left(E_1>E_2\right)$ are connected in potentiometer circuit, so as to assist each other. The null point obtained is at 490 cm from the higher potential end. When cell $E_2$ is connected, so as to oppose cell $E_1$, the null point is obtained at 90 cm from the same end. The ratio of the emfs of two cells $\left(\frac{E_1}{E_2}\right)$ is

Accepted Answer

The emf of a cell in a potentiometer is
$$E=\frac{V}{L} I$$
where, $I=$ length of wire at null point,
$V=$ voltage of source
and $L=$ total length of potentiometer wire.
$	herefore \quad E \propto I$
When two cell of emf $E_1$ and $E_2\left(E_1>E_2ight)$ are connected to assist each other, then
$$E_1+E_2=490\quad 	ext{... (i)}$$
When the cell of emf $E_2$ is connected, so as to oppose cell $E_1$, then
$$E_1-E_2=90\quad 	ext{.... (ii)}$$
From Eqs. (i) and (ii), we get
$$\begin{array}{ll}
& E_1=290 \mathrm{~V} 	ext { and } E_2=200 \mathrm{~V} \
	herefore & \frac{E_1}{E_2}=\frac{290}{200}=\frac{29}{20}=1.45
\end{array}$$

Answer

0.689

Answer

5.33

Answer

1.45

Answer

0.182

Current Electricity

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