Q133 / 180 · KCET 2018 · Physics

A \( 100 \mathrm{~W} \) bulb is connected to an \( \mathrm{AC} \) source of \( 220 \mathrm{~V}, 50 \mathrm{~Hz} \). Then the current flowing through
the bulb is

  • Correct answer
    A
    \( \frac{5}{11} A \)
  • B
    \( \frac{1}{2} A \)
  • C
    \( 2 \mathrm{~A} \)
  • D
    \( \frac{3}{4} A \)

Solution

Given, power of bulb \( =100 \mathrm{~W} \); voltage of source \( =200 \mathrm{~V} \)
Now, \( P=I V \Rightarrow I=\frac{P}{V}=\frac{100}{220}=\frac{5}{11} A \)
Therefore, current flowing through the bulb is \( \frac{5}{11} A \)

Hence, the correct answer is option: 'A'.

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