Question

A single phase thyristor controlled bridge rectifier is supplying a dc load of 1 kW. A 1.5 kVA isolated transformer with source side voltage rating of 120 V, 50 Hz is used. Its total leakage reactance is 8% based on its rating. The source voltage of nominally 115 V is in range of 10%. If the load current is constant, then the minimum turns ratio of the transformer, if the dc load voltage is to be regulated at constant value of 100 V will be ______ . (Answer up to 3 decimal places)

Answer 1

Base impedance of isolation transformer

$Z_b=\frac{V^2}{VA}=\frac{(120{)}^2}{1500}=9.6\Omega$

$X_{L-S}=\frac{8}{100}\times 9.6=0.768\Omega$

$\omega L_s=0.768\Omega$

$L_s=\frac{0.768}{2\pi \times 50}=2.445mH$

If 'a' is the turn ratio, $a=\frac{V_p}{V_s}$

$P_0=100\times I_0=1000$

So, $I_0=10A$

For minimum value of turn ratio,

$\alpha =0$

$V_p=115\times 0.9=103.5V$

$V_0=\frac{2\sqrt{2}}{\pi }{V}_s-\Delta V_0=\frac{0.9V_p}{a}-\left(\frac{4f{L}_s{I}_0}{a^2}\right)$

$=\frac{0.9\times 103.5}{a}-\frac{4\times 50\times 2.445\times {10}^{-3}\times 10}{a^2}=100$

$\frac{93.15}{a}-\frac{4.89}{a^2}=100$

$100a^2-93.15a+4.89=0$

$a=0.876,0.0558$

So, turn ratio, a = 0.876