The resistance of any bulb=V2/P

Current through ammeter at shown position=3i1
current in any 1 bulb=i1=RV=V2/PV=VP=220100
Through ammeter, ⇒inet=3i1=220300=1.35A
JEE Main 2014 — Physics Electromagnetism

Four bulbs B1, B2, B3 and B4 of 100W each are connected to 220V main as shown in the figure. The reading in an ideal ammeter will be
Held on 19 Apr 2014 · Verified 6 Jul 2026.
0.90A
1.35A
0.45A
1.80A
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