Total initial resistance

= R_G + R₁ = (50 + 2950) $\Omega$ = 3000 $\Omega$

$\varepsilon = 3V$

$\therefore$ Current = ${{3V} \over {3000\Omega }} = 1 \times {10^{ - 3}}mA$

If the deflection has to be reduced to 20 divisions, current i = 1 mA $\times$ ${2 \over 3}$ as the full deflection scale for 1 mA = 30 divisions.

$3V = 3000\Omega \times 1mA = x\Omega \times {2 \over 3}mA$

$\Rightarrow x = 3000 \times 1 \times {3 \over 2} = 4500\Omega$

But the galvanometer resistance = 50 $\Omega$

Therefore the resistance to be added

= (4500 – 50)$\Omega$ = 4450 $\Omega$.