MHT CET · Physics · Alternating Current
A capacitor \(50 \mu\) Fis connected to a.c. source \(e=220 \sin (50 t)(e\) in volt ,\(t\) in second). The value of peak current is
- A \(\frac{0.55}{\sqrt{2}} A\)
- B \(\sqrt{2}\)
- C \(\frac{\sqrt{2}}{0.55} A\)
- D \(0.55 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(\frac{0.55}{\sqrt{2}} A\)
Step-by-step Solution
Detailed explanation
Power source \(V=220 \sin 50 t\)
\(\therefore\) Peak voltage \(V_0=220\) volt
So, rms voltage \(\quad V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=\frac{220}{\sqrt{2}}\) volt
Also \(\omega=50\)
\(\therefore\) capacitive reactance \(X_C=\frac{1}{\omega C}=\frac{1}{50 \times 50 \times 10^{-6}}=400 \Omega\)
value of rmscurrent \(I_{r m s}=\frac{V_{r m s}}{X_C}=\frac{220}{\sqrt{2} \times 400}=\frac{0.55}{\sqrt{2}}\) ampere
\(\therefore\) Peak voltage \(V_0=220\) volt
So, rms voltage \(\quad V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=\frac{220}{\sqrt{2}}\) volt
Also \(\omega=50\)
\(\therefore\) capacitive reactance \(X_C=\frac{1}{\omega C}=\frac{1}{50 \times 50 \times 10^{-6}}=400 \Omega\)
value of rmscurrent \(I_{r m s}=\frac{V_{r m s}}{X_C}=\frac{220}{\sqrt{2} \times 400}=\frac{0.55}{\sqrt{2}}\) ampere
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