The galvanometer deflection, when key $\mathrm{K}_{1}$ is closed but $\mathrm{K}_{2}$ is open, equals $\theta_{0}$ (see figure). On closing $K_{2}$ also and adjusting $\mathrm{R}_{2}$ 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]:

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Current Electricity

The galvanometer deflection, when key

K_{1}

is closed but

K_{2}

is open, equals

θ_{0}

(see figure). On closing

K_{2}

also and adjusting

R_{2}

5Ω

, the deflection in galvanometer becomes

\frac{θ _{0}}{5}

. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

Answer: (B)

With

K_{1}

closed and

K_{2}

open the galvanometer carries the full current

i_{1} = \frac{E}{220 + G}

, where

220 Ω

is the fixed series resistance and

G

the galvanometer resistance; this produces the deflection

θ_{0}

. Closing

K_{2}

puts the

5 Ω

resistor in parallel with the galvanometer, so the current splits and only the fraction

\frac{5}{G + 5}

of the line current reaches the coil. Working out the galvanometer current in this second state and demanding that it be one-fifth of

i_{1}

, since the deflection drops to

\frac{θ _{0}}{5}

, gives

\frac{5 ( 220 + G )}{220 ( G + 5 ) + 5 G} = \frac{1}{5}

. Cross-multiplying,

25 (220 + G) = 225 G + 1100

, which simplifies to

4400 = 200 G

, so

G = 22 Ω

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option