< 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 \(\frac{2.4}{4.5}=0.53\ Amp\).

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

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

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

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

Fig. 28(b)

Fig. 28(c)

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

To find the R_th the circuit diagram may be redrawn as

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

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