JEE Mains · Physics · STD 12 - 3. current electricity
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 ............... \(A\)
- A \(0.45\)
- B \(0.90\)
- C \(1.35\)
- D \(1.80\)
Answer & Solution
Correct Answer
(C) \(1.35\)
Step-by-step Solution
Detailed explanation
Current in each bulb \(=\frac{\text { Power }}{\text { Voltage }}\) \(=\frac{100}{220}=0.45\, \mathrm{A}\) Current through ammeter \(=0.45 \times 3=1.35\, \mathrm{A}\)
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