Explanation:
Resistance of any bulb is
\(R = \frac{{{V^2}}}{P} = \frac{{{{(220)}^2}}}{{100}} = 484\,\Omega \)
Net resistance of the circuit is
\({R_{eq}} = \frac{R}{4} = \frac{{484}}{4} = 121\,\Omega \)
\(V = i\,{R_{eq}} \Rightarrow \,\,220 = i \times 121\)
\(i = 1.81\,\,Amp\)
Current in each branch \( = \frac{{1.81}}{4}amp = 0.45{\kern 1pt} \,\,amp\)
Reading of ammeter \( = 3 \times 0.45 = 1.35\,\,amp\)