When an ac generator of 120,V is connected in series with a capacitor and a resistor of 30,Ω , circ

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7.Alternating Current

easy

When an ac generator of $120\,V$ is connected in series with a capacitor and a resistor of $30\,\Omega $, the circuit carries a current $1·5\,A$. The potential difference across the capacitor will be.....$V$

A

$1.11$

B

$111$

C

$0$

D

$220$

Solution

$\mathrm{V}^{2} =\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{C}}^{2}$

$\mathrm{V}_{\mathrm{C}} =\sqrt{\mathrm{V}^{2}-\mathrm{V}_{\mathrm{R}}^{2}}=\sqrt{(120)^{2}-(1.5 \times 30)^{2}} $

$=111 \mathrm{\,volt} $

Standard 12

Physics

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