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 [tex]v_p=120volt[/tex]
(i) We know that peak voltage is give by [tex]v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt[/tex]
(ii) We know that for transformer [tex]\frac{v_p}{v_s}=\frac{n_p}{n_s}[/tex]
So [tex]\frac{169.08}{v_s}=\frac{10}{1}[/tex]
[tex]v_s=16.90volt[/tex]
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 [tex]v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt[/tex]
(v) Now dc current is given by [tex]i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A[/tex]
