A) 200 V, 1A
B) 800 V, 2A
C) 100 V, 2A
D) 220 V, 2.2A
Correct Answer: D
Solution :
Key Idea: Voltage across inductance and capacitance is equal, hence cancels out. In the series L-C-R circuit \[U=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] (voltage across resistance) is in the phase with current (i), while \[=9\times {{10}^{9}}\times 1\times {{10}^{-6}}\times {{10}^{-3}}\left( \frac{1}{1}-\frac{1}{10} \right)\] (voltage across inductance) leads i by \[=8.1J\], while Ye lags behind i by \[KE=\frac{1}{2}m{{v}^{2}}=8.1J\]. Resultant emf E is \[\Rightarrow \] Since, \[v=\sqrt{\frac{2\times 8.1}{2\times {{10}^{-3}}}}=90m/s\] \[F=eE\] Reading of ammeter, \[=\frac{force}{mass}=\frac{eE}{m}\]
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