A galvanometer G of $$2\Omega$$ resistance is connected in the given circuit. The ratio of charge stored in $$C_1$$ and $$C_2$$ is

step 1: steady state idea

capacitors act as open circuits
so only resistors + galvanometer (2 Ω) matter

step 2: find equivalent resistance

the network reduces to:

4 Ω + 2 Ω + 6 Ω = 12 Ω

step 3: current

$$I=\frac{6}{12}=0.5\text{ A}$$

step 4: potential distribution

drop across 4 Ω:

$$V=0.5\times4=2\text{ V}$$

drop across 2 Ω:

$$V=0.5\times2=1\text{ V}$$

drop across 6 Ω:

$$V=0.5\times6=3\text{ V}$$

step 5: capacitor voltages

from node potentials:

• voltage across $$C_1$$​ = 3 V
• voltage across $$C_2$$ = 4 V

step 6: charges

$$q_2=C_2V_2=6\mu F\times4=24\mu C$$

$$q_1=C_1V_1=4\mu F\times3=12\mu C$$