A $100 \mu \mathrm{F}$ capacitor is connected to a $100 \mathrm{~V}$, $50 \mathrm{~Hz}$ AC supply. The rms…

A $100 \mu \mathrm{F}$ capacitor is connected to a $100 \mathrm{~V}$, $50 \mathrm{~Hz}$ AC supply. The rms value of the current is
  1. $3.14 \mathrm{~A}$
  2. $4.75 \mathrm{~A}$
  3. $2.33 \mathrm{~A}$
  4. $5.5 \mathrm{~A}$

Solution

Given, $V_{\mathrm{rms}}=100 \mathrm{~V}, f=50 \mathrm{~Hz}$ $C_1=100 \mu \mathrm{F}=100 \times 10^{-6} \mathrm{~F}$ Now, reactance of capacitor, $X_C=\frac{1}{C \omega}=\frac{1}{C \times 2 \pi f}$ $\Rightarrow \quad X_C=\frac{1}{100 \times 10^{-6} \times 2 \pi \times 50}=\frac{100}{\pi} \Omega$ r. m. s value of current in circuit will be $I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}$ $=\frac{100}{\frac{100}{\pi}}=\pi \mathrm{A}=3.14 \mathrm{~A}$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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