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  • The equation of current in a purely inductive circuit is <img alt="5 \sin \left(49 \pi t-30^{\circ}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?5%20%5Csin%20%5Cleft%2849%20%5Cpi%20

The equation of current in a purely inductive circuit is 5 \sin \left(49 \pi t-30^{\circ}\right). If the inductance is 30 \mathrm{mH} then the equation for the voltage across the inductor, will be :
\left\{\right.$ Let $\left.\pi=\frac{22}{7}\right\}
 

Option: 1

\mathrm{1.47 \sin \left(49 \pi \mathrm{t}-30^{\circ}\right)}
 


Option: 2

\mathrm{1.47 \sin \left(49 \pi t+60^{\circ}\right)}


Option: 3

\mathrm{23.1 \sin \left(49 \pi t-30^{\circ}\right)}


Option: 4

\mathrm{23.1 \sin \left(49 \pi t+60^{\circ}\right)}


Answers (1)

best_answer

\mathrm{I_L =5 \sin \left(49 \pi t-30^{\circ}\right) }

\mathrm{L =30 \mathrm{mH} }

\mathrm{X_L =\omega L=2 \pi f L }

        \mathrm{=49 \pi \times 30 \times 10^{-3} }

        \mathrm{=49 \times \frac{22}{7} \times 30 \times 10^{-3} }

\mathrm{X_L =210 \times 22 \times 10^{-3} }

\mathrm{X_L =462 \times 10^{-2}}

\mathrm{V_L=\left(I_{L_0}\right)X_L}

\mathrm{V_L=(23.1) \sin \left(49 \pi t-30^{\circ}+90\degree\right) }

Voltage is ahead of current by \mathrm{90^{\circ}} in pure inductive ckt

\mathrm{ V_L=23.1 \sin \left(49 \pi t+60^{\circ}\right) }

Hence 4 is correct option.



 

Posted by

Gaurav

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