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GUJCET · Physics · Alternating Current
A \(50 \mu \mathrm{~F}\) capacitor is connected to a \(110 \mathrm{~V}, 60 \mathrm{~Hz}\) AC supply. Determine the rms value of the current in the circuit.
- A 2.5 A
- B 3.8 A
- C 5.2 A
- D 2.1 A
Answer & Solution
Correct Answer
(D) 2.1 A
Step-by-step Solution
Detailed explanation
(D) 2.1 A
Capacitive Reactance
\(X_{C}=\frac{1}{\omega C}\)
\(X_{C}=\frac{1}{2 \pi \nu C}\)
\(X_{C}=\frac{1}{2 \pi \times 60 \times 50 \times 10^{-6}}\)
\(X_{C}=\frac{1}{60 \times 3.14 \times 10^{-4}}\)
\(\mathrm{X}_{\mathrm{C}} \approx 53 \Omega\)
c) rms current in the circuit
\(
\begin{aligned}
& I=\frac{V}{X_{C}} \\
& I=\frac{110}{53} \\
& I=2.1 \mathrm{~A}
\end{aligned}
\)
Capacitive Reactance
\(X_{C}=\frac{1}{\omega C}\)
\(X_{C}=\frac{1}{2 \pi \nu C}\)
\(X_{C}=\frac{1}{2 \pi \times 60 \times 50 \times 10^{-6}}\)
\(X_{C}=\frac{1}{60 \times 3.14 \times 10^{-4}}\)
\(\mathrm{X}_{\mathrm{C}} \approx 53 \Omega\)
c) rms current in the circuit
\(
\begin{aligned}
& I=\frac{V}{X_{C}} \\
& I=\frac{110}{53} \\
& I=2.1 \mathrm{~A}
\end{aligned}
\)
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