InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
An laternativing voltage of 350 V, 60 Hz is applied on a full wave rectifier. The internal resistance of each diode is 200Omega. If R_(L)=5kOmega, then find (i) the peak value of output current. (ii) the value of output direct current. (iii) theoutput dc power. (iv) the rms value of output current. (v) the efficiency of rectifier. (vi) the value of peak inverse voltage (P.I.V.). |
|
Answer» <P> Solution :(i) `I_("PEAK")=I_("rms")xxsqrt2=(V_("rms")xxsqrt2)/((R_(L)+2r_(p)))``or I_(0)=(350xxsqrt5)/(5400)=0.092A` (II) `I_(DC)=(2I_(0))/(pi)=(2xx0.092)/(3.14)=0.058A` (iii) `p_(DC)=I_(DC)^(2)xxR_(L)=(0.058)^(2)xx(5000)=17W` (iv) `I_("rms")=(I_(0))/(sqrt2)=(0.092)/(1.41)=0.065A` (v) Efficiency of reaction `eta=(81.6)/(1+(r_(p))/(R_(L)))` `or eta=(81.6)/(1+(200)/(5000))=(81.6xx25)/(26) or eta=78%` (VI) `P.I.V.=2E_(0)` `or P.I.V.=2sqrt2E_("rms")=2sqrt2=350` `or P.I.V.=1000V`. |
|