Question 363: Current Electricity

Question

The galvanometer deflection, when key K
1
is closed but K
2
is open, equals $	heta $
0
(see figure). On closing K
2
also and adjusting R
2
to 5$\Omega $, the deflection in galvanometer becomes ${{{	heta _0}} \over 5}.$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :
![](https://imagex.cdn.examgoal.net/YvafhLee5rnnELJ8z/0cL9UEVn22HZf90zKpslapPiiecBs/Vu3wEHKZ54jKwYBRwxkYu8/image.png)

Accepted Answer

case I :
i
g
= ${E \over {220 + {R_g}}}$ = C$	heta $
0
. . .(i)
Case II :
i
g
= $\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} ight)$ $ 	imes $ ${5 \over {\left( {{R_g} + 5} ight)}}$ = ${{C{	heta _0}} \over 5}$       . . .(ii)
$ \Rightarrow $  ${{5E} \over {225{R_g} + 1100}}$ = ${{C{	heta _0}} \over 5}$       . . .(ii)
${E \over {220 + {R_g}}} = C	heta $       . . .(i)
$ \Rightarrow $  ${{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5$
$ \Rightarrow $  5500 + 25R
g
= 225R
g
+ 1100
200R
g
= 4400
R
g
= 22$\Omega $

Answer

5 $\Omega $

Answer

25 $\Omega $

Answer

12 $\Omega $

Answer

22 $\Omega $

Current Electricity

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