7.Alternating Current
hard

$0.7\,H$ ઇન્ડકટર અને $220\,\Omega$ અવરોધને $220\,V , 50\,Hz$ ના $a.c.$ ઉદગમ સાથે જોડવામાં આવે છે,તો પ્રવાહ એ વૉલ્ટેજમાં કેટલી કળામાં પાછળ અને પ્રવાહ શોધો.

A

$30^o,\,\,1\,A$

B

$45^o,\,\,0.5\,A$

C

$60^o,\,\,1.5\,A$

D

આપેલ પૈકી એક પણ નહીં

(AIIMS-2010)

Solution

$\mathrm{L}=0.7 \mathrm{\,H}, \mathrm{R}=220 \Omega, \mathrm{E}_{0}=220 \mathrm{\,V}, v=50 \mathrm{\,Hz}$

This is an $L – R$ circuit Phase difference,

$\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}=\frac{\omega \mathrm{L}}{\mathrm{R}}=\frac{2 \pi v \mathrm{L}}{\mathrm{R}}$

$\left[\mathrm{X}_{\mathrm{L}}=2 \pi v \mathrm{L}=2 \times \frac{22}{7} \times 50 \times 0.7=220 \Omega\right]$

$=\frac{220}{220}=1$ or, $\phi=45^{\circ}$

Wattless component of current

$=I_{0} \sin \phi=\frac{I_{0}}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{E_{0}}{Z}$

$=\frac{1}{\sqrt{2}} \frac{220}{\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}}=\frac{1}{\sqrt{2}} \cdot \frac{220}{\sqrt{220^{2}+220^{2}}}$

$=\frac{1}{2}=0.5 \mathrm{\,A}$

Standard 12
Physics

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