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A $60 \mu \mathrm{F}$ capacitor is connected to a $110 \mathrm{~V}, 60 \mathrm{~Hz}$ ac supply. Determine the rms value of the current in the circuit.
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Verified Answer
Given: capacitance $\mathrm{C}=60 \mu \mathrm{F}$
voltage $\mathrm{E}_{\mathrm{v}}=110 \mathrm{~V}$ frequency $v=60 \mathrm{~Hz}$
To find: rms current
$$
\begin{aligned}
&\text { Formula: } \mathrm{I}_{\mathrm{v}}=\frac{\mathrm{E}_{\mathrm{v}}}{\mathrm{X}_{\mathrm{c}}}=\frac{\mathrm{E}_{\mathrm{v}}}{1 / \omega \mathrm{C}}=\mathrm{E}_{\mathrm{v}} \times \omega \mathrm{C} \\
&=\mathrm{E}_{\mathrm{v}} \times 2 \pi v \mathrm{C}=110 \times 2 \times 3.14 \times 60 \times 60 \times 10^{-6} \\
&=1.1 \times 6.28 \times 36 \times 10^{-2}=2.49 \mathrm{~A}
\end{aligned}
$$
voltage $\mathrm{E}_{\mathrm{v}}=110 \mathrm{~V}$ frequency $v=60 \mathrm{~Hz}$
To find: rms current
$$
\begin{aligned}
&\text { Formula: } \mathrm{I}_{\mathrm{v}}=\frac{\mathrm{E}_{\mathrm{v}}}{\mathrm{X}_{\mathrm{c}}}=\frac{\mathrm{E}_{\mathrm{v}}}{1 / \omega \mathrm{C}}=\mathrm{E}_{\mathrm{v}} \times \omega \mathrm{C} \\
&=\mathrm{E}_{\mathrm{v}} \times 2 \pi v \mathrm{C}=110 \times 2 \times 3.14 \times 60 \times 60 \times 10^{-6} \\
&=1.1 \times 6.28 \times 36 \times 10^{-2}=2.49 \mathrm{~A}
\end{aligned}
$$
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