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7.Alternating Current
hard
$\mathrm{AC}$ emf $\mathrm{E}=110 \sqrt{2} \sin 100 \mathrm{t}$ volt, એક $2 \mu \mathrm{F}$ ના કેપેસીટરને $\mathrm{E}=110 \sqrt{2} \sin 100 \mathrm{t}$ જેટલો ઉલટસૂલટ ( $\mathrm{AC}) \mathrm{emf}$ લગાડવામાં આવે તો પરિપથમાં પ્રવાહનું $rms$. . . . . . .$\mathrm{mA}$ થાય.
A
$22$
B
$20$
C
$25$
D
$30$
(JEE MAIN-2024)
Solution
$C=2 \mu f ; \quad E=110 \sqrt{2} \sin (100 \mathrm{t})$
$\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega c}=\frac{1}{100 \times 2 \times 10^6}$
$=\frac{10000}{2}=5000 \Omega$
$\mathrm{i}_0=\frac{110 \sqrt{2}}{5000}$
$\mathrm{i}_{\mathrm{rms}}=\frac{110 \sqrt{2}}{5000 \sqrt{2}}$
$=\frac{110}{5} \mathrm{~mA}$
$=22 \mathrm{~mA}$
Standard 12
Physics
