| Direction: Q.36 to Q.40 |
| In an AC circuit, values of voltage and current change every instant. Therefore, power of an AC circuit at any instant is the product of instantaneous voltage (E) and instantaneous current (I). The average power supplied to a pure resistance R over a complete cycle as AC is \[P={{E}_{v}}{{I}_{v}}\]. When circuit is inductive, average power per cycle is \[{{E}_{v}}{{I}_{v}}\,\cos \phi \]. |
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| In an AC circuit, 600 mH inductor and a \[50\,\mu F\]capacitor are connected in series with \[10\,\Omega \] resistance. The AC supply to the circuit is 230 V, 60 Hz. |
| Based on the above information, answer the following questions. |
A) 10.42 W
B) 15.25 W
C) 17.42 W
D) 13.45 W
Correct Answer: C
Solution :
(c) 17.42 W Average power transferred per cycle to resistance is \[{{P}_{v}}=l_{v}^{2}R\] As \[{{X}_{L}}=\omega L=2\pi vL\] \[=2\times \frac{22}{7}\times 60\times 0.6=226.28\,\Omega \] \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi vC}=\frac{1}{2\times 22/7\times 60\times 50\times {{10}^{-6}}}\] \[=53.03\,\Omega \] \[Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}\] \[=\sqrt{{{\left( 10 \right)}^{2}}+{{\left( 226.28-53.03 \right)}^{2}}}=173.53\,\Omega \] \[{{l}_{v}}=\frac{{{E}_{v}}}{Z}=\frac{230}{173.53}=1.32\,A\] \[{{P}_{v}}=l_{v}^{2}R={{(1.32)}^{2}}\times 10=17.42\,\,W\]
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