Thank you for reporting, we will resolve it shortly
Q.
Four bulbs $B_1,\, B_2,\, B_3$ and $B_4$ of $100\, W$ each are connected to $220\, V$ main as shown in the figure. The reading in an ideal ammeter will be :
Resistance of any bull
$R =\frac{ V ^{2}}{ P }=\frac{(220)^{2}}{100}=484 \,\Omega$
Net resistance of the cks
$R _{ eq }=\frac{ R }{4}=\frac{484}{4}=121 \,\Omega$
$v = i R _{ eq }$
$220 \times i \times 121$
$i =1.81$ Amp. (total current supplied by the battery)
Current in each branch in $i =\frac{1.81}{4}$ amp.
$=0.45$ amp.
Reading of ammeter $=3 \times 0.45=1.3575$ amp.
