Calculate |
(i) the impedance of the circuit and |
(ii) the peak value of the current flowing in the circuit |
(iii) the power factor of this circuit and compare this value with the one at its resonant frequency. |
Answer:
(i) \[100\sqrt{2}\] (ii) 2.2 A (iii) \[1/\sqrt{2}\] Given, \[R=100\Omega \] \[L=\frac{2}{\pi }H,\,\,C=\frac{100}{\pi }\mu F\] \[{{V}_{ms}}=220\,V,\,\,f=50Hz\] (i) \[Z=\sqrt{{{R}^{2}}+{{({{X}_{c}}-{{X}_{L}})}^{2}}}\] \[{{X}_{c}}=\frac{1}{2\pi fC}=\frac{1}{2\pi \times 50\times \frac{100}{\pi }\times {{10}^{-\,6}}}=100\Omega \] \[{{X}_{L}}=2\pi fL=2\pi \times 50\times \frac{2}{\pi }=200\Omega \] \[\therefore \] \[Z=\sqrt{{{100}^{2}}+{{(100-200)}^{2}}}=100\sqrt{2}\] (ii) \[{{V}_{m}}=\sqrt{2}\,{{V}_{rms}}=220\sqrt{2}\,V\] \[{{i}_{m}}=\frac{{{V}_{m}}}{z}=\frac{\sqrt{2}\times 220}{100\sqrt{2}}=2.2\,A\] (ii) Power factor \[=\cos \,\phi \] Where, \[\phi ={{\tan }^{-1}}\frac{{{X}_{c}}-{{X}_{L}}}{R}={{\tan }^{-1}}(-1)=-\,45{}^\circ \] \[\therefore \] Power factor \[=\cos \,(-\,45{}^\circ )=\frac{1}{\sqrt{2}}=0.707\] At resonance, power factor \[=1\]
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