Question 588: Current Electricity

Question

In a meter bridge experiment, the balance point is obtained at length $$\ell_1 \mathrm{~cm}$$ from left end when resistances in the left gap and right gap are $$5 \Omega$$ and $$R \Omega$$ respectively. When the resistance $$R$$ is shunted with equal resistance, the new balance point is at $$1.6 \ell_1$$. The resitance $$R$$ in ohm is

Accepted Answer

$$\begin{aligned}
& \frac{5}{R}=\frac{\ell_1}{100 \ell_1} \quad 	ext{.......(1)}\
& \frac{5 	imes 2}{R}=\frac{1.6 \ell_1}{100-1.6 \ell} \
& 	herefore \quad \frac{5}{R}=\frac{0.8 \ell_1}{100-1.6 \ell_1} \quad 	ext{.......(2)}
\end{aligned}$$
Equating (1) and (2) and solving we get $$\ell_1=25 \mathrm{~cm}$$ Putting this in Eq.(1), we get $$\mathrm{R}=15 \Omega$$

Answer

25

Answer

15

Answer

10

Answer

20

Current Electricity

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