A) 109mA
B) 126mA
C) 150mA
D) 164mA
Correct Answer: D
Solution :
Given that \[R=200\,\Omega ,\,C=15\mu F=15\times {{10}^{-6}}F\] \[L=230\,mH=230\times {{10}^{-3}}H\] and \[\varepsilon =36.0\sin 120\,\pi t\] This is comparing with \[\varepsilon ={{\varepsilon }_{0}}\sin \omega \,t\] \[{{\varepsilon }_{0}}=36\] \[\omega =120\pi \] \[{{X}_{L}}=\omega L\] \[=120\pi \times 230\times {{10}^{-3}}\] \[=86.664\approx 86.66\] \[{{X}_{C}}=\frac{1}{\omega c}=\frac{1}{120\pi \times 15\times {{10}^{-6}}}\] \[=\frac{{{10}^{8}}}{120\times 314\times 15}\] \[=176.93\,\Omega \] \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\] \[=\sqrt{40000+8148.67}\] \[=\sqrt{48148.67}\] \[=219.42\,\Omega \] \[i=\frac{{{\varepsilon }_{0}}}{2}=\frac{36}{219.42}\] =0.164 A = 164 mA
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