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, :
![](https://imagex.cdn.examgoal.net/yaYFpvE7Odr4MpHl5/KZYYPl0eQVGv3ClDPEj3CfCUCQxzg/2RaT3JLFUhpnojwgKT6lxG/image.png)

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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