Question

Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier. Rload is 50 ohms. Round off to the nearest tenths place. i. Primary or Input Peak Voltage ii. Secondary Peak Voltage iii. Peak Voltage out of the rectifier iv. DC Voltage out of the rectifier v. DC Current into the load

Answer 1

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage $v_p=120volt$

(i) We know that peak voltage is give by $v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt$

(ii) We know that for transformer $\frac{v_p}{v_s}=\frac{n_p}{n_s}$

So $\frac{169.08}{v_s}=\frac{10}{1}$

$v_s=16.90volt$

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by $v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt$

(v) Now dc current is given by $i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A$