< style="display: block; margin-left: auto; margin-right: auto;" src="http://mindstudy.in/storage/ee-posts/June2022/Fig. 28(a).png" alt="" width="293" height="223" />Fig.28(a)

Total resistance of parallel pair $\frac{9\times 9}{9+9}=4.5\mathrm{\Omega }$

Source current is $\frac{2.4}{4.5}=0.53Amp$.

Current the branch BAD is $\frac{2.4}{4.5\times 2}=\frac{1.2}{4.5}=0.27Amp$.

Current through the branch BCD is $\frac{2.4}{4.5\times 2}=\frac{1.2}{4.5}=0.27Amp$.

Voltage drop across BA is $\frac{4\times 1.2}{4.5}=1.07V$

Voltage drop across BC is $\frac{3\times 1.2}{4.5}=0.8V$

Fig. 28(b)

Fig. 28(c)

Hence voltage across AC is $V_{o.c}=1.07-0.8=0.27V$

To find the R_th the circuit diagram may be redrawn as

Hence R_th =$(8+2.2+2)=12.2\mathrm{\Omega }$

The current through the galvanometer is $\frac{0.27}{12.2}=0.02Amp$.