\( V = 200\sin(400\,t) \) volt a.c. source is connected to a series combination of an inductance of 20 mH and a resistance of 6 \( \Omega \). What will be the rms current in the a.c. circuit?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 14.14\,\mathrm{A} \)
Comparing \( V = V_{0}\sin\omega t \) with the given source, \( V_{0} = 200\,\mathrm{V} \) and \( \omega = 400\,\mathrm{rad\,s^{-1}} \). In a series LR circuit the current is set by the impedance \( Z = \sqrt{R^{2} + X_{L}^{2}} \).
\( \displaystyle X_{L} = \omega L = 400 \times 20\times10^{-3} = 8\,\Omega \)
\( \displaystyle Z = \sqrt{6^{2} + 8^{2}} = 10\,\Omega \)
\( \displaystyle I_{0} = \frac{V_{0}}{Z} = \frac{200}{10} = 20\,\mathrm{A} \)
The rms current is the peak value divided by \( \sqrt{2} \).
\( \displaystyle I_{\mathrm{rms}} = \frac{I_{0}}{\sqrt{2}} = \frac{20}{1.414} = 14.14\,\mathrm{A} \)