Question 289: Current Electricity

Question

The galvanometer deflection, when key K1 is closed but K2 is open, equals $$	heta $$0 (see figure). On closing K2 also and adjusting R2 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] :

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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