Question 369: Current Electricity

Question

The resistance of the meter bridge AB in given figure is 4 $$\Omega $$. With a cell of emf $$\varepsilon $$ = 0.5 V and rheostat
resistance R
h
= 2 $$\Omega $$ the null point is obtained at some point J. When the cell is replaced by another one of emf $$\varepsilon $$ = $$\varepsilon $$
2
the same null point J is found for R
h
= 6 $$\Omega $$. The emf $$\varepsilon $$
2
is, :

Accepted Answer

Potential gradient with R
h
= 2$$\Omega $$
is $$\left( {{6 \over {2 + 4}}} ight) 	imes {4 \over L} = {{dV} \over {dL}};$$   L $$=$$ 100 cm
Let null point be at $$\ell $$ cm
thus $$\varepsilon $$
1
$$=$$ 0.5V $$=$$ $$\left( {{6 \over {2 + 4}}} ight) 	imes {4 \over L} 	imes \ell $$      . . .(1)
Now with R
h
$$=$$ 6$$\Omega $$ new potential gradient is
$$\left( {{6 \over {4 + 6}}} ight) 	imes {4 \over L}$$  and at null point
$$\left( {{6 \over {4 + 6}}} ight)\left( {{4 \over L}} ight) 	imes \ell  = {\varepsilon _2}$$                          . . .(2)
dividing equation (1) by (2) we get
$${{0.5} \over {{\varepsilon _2}}} = {{10} \over 6}$$  thus  $${\varepsilon _2} = 0.3$$

Answer

0.3 V

Answer

0.6 V

Answer

0.5 V

Answer

0.4 V

Current Electricity

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