Question

The galvanometer deflection, when key k1 is closed but k2 is open, equals theetha

Answer 1

Case 1

ig=\frac{E}{220+R_{g}}=c\Theta _{0}\, \, \, \, \, \, (1)

Case 2

ig=\left ( \frac{E}{220+\frac{5R_{g}}{5+R_{g}}} \right )\frac{5}{R_{g}+5}=\frac{c\Theta _{0}}{5}\, \, \, \, \, \, (2)

\left ( 1 \right )/(2)

\Rightarrow \frac{225R_{g}+1100}{1100+5R_{g}}=5

\Rightarrow R_{g}=22\Omega

Option 1)

5 \Omega

Option 2)

12 \Omega

Option 3)

25 \Omega

Option 4)

22 \Omega

Answer 2

Answer:

Explanation:

Missing question :- The galvanometer deflection , when key K1 is closed but K2 is open, equals \theta _{0} (see figure). On closing K2 also and adjusting R2 to 5 \Omega , the deflection in galvanometer becomes \frac{\theta _{0}}{5} . The resistance of the galvanometer is then , given by [ Neglect the internal resistance of battery ] :

Answer :-

Equivalent Resistance = G(s) /G + s

For Case 1

ig = E / 220 + R(g) = c Ф(o) --- (1)

For Case 2

ig = ( E / 220 + (5 Rg/ 5 + Rg) ) 5 / 5 + Rg =  c Ф(o)/5 .... (2)

1 divide 2 we get,

(225 Rg + 1100)/ (1100 + 5 Rg) = 5

Rg = 22 Ohms