NEET · Physics · STD 12 -7. Alternating current
A \(10 \mu \mathrm{F}\) capacitor is connected to a \(210 \mathrm{~V}, 50 \mathrm{~Hz}\) source as shown in figure. The peak current in the circuit is nearly \((\pi=3.14)\) :
- A \(0.93 \mathrm{~A}\)
- B \(1.20 \mathrm{~A}\)
- C \(0.35 \mathrm{~A}\)
- D \(0.58 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(0.93 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\text { Capacitive Reactance } X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}}\) \(=\frac{1000}{3.14}\)
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