KCET · Physics · Alternating Current
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
- A \( \frac{5}{11} A \)
- B \( \frac{1}{2} A \)
- C \( 2 \mathrm{~A} \)
- D \( \frac{3}{4} A \)
Answer & Solution
Correct Answer
(A) \( \frac{5}{11} A \)
Step-by-step Solution
Detailed explanation
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 \( \frac{5}{11} A \)
Now, \( P=I V \Rightarrow I=\frac{P}{V}=\frac{100}{220}=\frac{5}{11} A \)
Therefore, current flowing through the bulb is \( \frac{5}{11} A \)
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