A) 40%
B) 20%
C) 30%
D) 15%
Correct Answer: A
Solution :
| \[{{I}_{m}}=\frac{{{V}_{m}}}{{{R}_{f}}+{{R}_{L}}}=\frac{25}{(10+1000)}=24.75\,mA\] |
| \[{{I}_{dc}}=\frac{{{I}_{m}}}{\pi }=\frac{24.75}{3.14}=7.87\,mA\] |
| \[{{I}_{rms}}=\frac{{{I}_{m}}}{2}=\frac{24.75}{2}=12.37\,mA\] |
| \[{{P}_{dc}}={{I}_{dc}}^{2}\times {{R}_{L}}={{(7.87\times {{10}^{-3}})}^{2}}\times {{10}^{3}}=61.9\,mW\] |
| \[{{P}_{ac}}={{I}_{rms}}^{2}({{R}_{f}}+{{R}_{L}})={{(12.37\times {{10}^{-3}})}^{2}}\times (10+1000)\]\[=154.54\,mW\] |
| Rectifier efficiency \[\eta =\frac{{{P}_{dc}}}{{{P}_{ac}}}\times 100=\frac{61.9}{154.54}\times 100=40.05%\] |
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