| Direction: Q.11 to Q.15 |
| The power averaged over one full cycle of AC is known as average power. It is also known as true power. \[{{P}_{av}}={{V}_{rms}}{{I}_{rms}}\cos \phi =\frac{{{V}_{0}}{{I}_{0}}}{2}\cos \phi \]. |
| Root mean square or simply rms watts refer to continuous power. |
| A circuit containing a 80 mH inductor and a \[60\text{ }\mu F\] capacitors in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. |
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| Read the above passage carefully and give the answer of the following questions: |
A) 15 A
B) 11.63 A
C) 17.65 A
D) 6.33 A
Correct Answer: B
Solution :
(b) 11.63 A Inductance, \[L=80\text{ }mH=80\times {{10}^{-3}}H\] Capacitance, \[C=60\,\mu F=60\times {{10}^{-6}}F,V=230\,V\] Frequency, \[v=50\text{ }Hz\] \[\omega =2\pi v=100\,\pi \,\,rad\,{{s}^{-1}}\] Peak voltage, \[{{V}_{0}}=V\sqrt{2}=230\,\sqrt{2}\,V\] Maximum current is given by, \[{{l}_{0}}=\frac{{{V}_{0}}}{\left( \omega L-\frac{1}{\omega C} \right)}\] \[{{l}_{0}}=\frac{230\sqrt{2}}{\left( 100\pi \times 80\times {{10}^{-3}}-\frac{1}{100\pi \times 60\times {{10}^{-6}}} \right)}\] \[{{l}_{0}}=\frac{230\sqrt{2}}{\left( 8\pi -\frac{1000}{6\pi } \right)}=-11.63\,A\] Amplitude of maximum current, \[{{\text{l}}_{0}}=11.63\text{ }A\]
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