Question 106: Current Electricity

Question

A resistance of 2 $$\Omega$$ is connected across one gap of a metre-bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 $$\Omega$$, is connected across the other gap. When these resistance are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is

Accepted Answer

Given, that x is greater than 2 $$\Omega$$, when the bridge is balanced then,
$$\frac{R}{l}=\frac{x}{100-l}$$ ..... (i)
(from metre bridge balanced bridge formula)
$$\Rightarrow 100R-Rl=lx$$
$$\Rightarrow 200-2l=lx~~(\because R=2\Omega)$$
$$l=\frac{200}{x+2}$$ ..... (ii)
Where, $$l$$ = length of segment of wire from one end where null point is obtained.
Now, these resistances are interchanged, the Jockey shifts 20 cm. So,
$$\frac{x}{l+20}=\frac{2}{80-l}$$ ...... (iii)
From equation (iii) we have,
$$80x=l(x+2)+40$$
$$\Rightarrow 80x=\left(\frac{200}{x+2}ight)(x+2)+40$$
$$\Rightarrow x=\frac{240}{80}=3\Omega$$
$$	herefore x=3\Omega$$

Answer

3 $$\Omega$$

Answer

4 $$\Omega$$

Answer

5 $$\Omega$$

Answer

6 $$\Omega$$

Current Electricity

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