Question

In a full wave junction diode rectifier the input ac has rms value of 20 V. the transformer used is a step up transformer having primary and secondary turn ration 1 : 2 The dc voltage in the rectified output is

• A. 12 V
• B. 24 V
• C. 36 V
• D. 42 V

Answer 1

Answer: (C)

36 V

Answer 2

: here, input$V_{rms} = 20V$

Peak value of input voltage

$V_{o} = \sqrt{2}V_{rms} = \sqrt{2} \times 20 = 28.28V$

Since the transformer is a step up transformer having transformer ratio $1:2$the maximum value of output voltage of the transformer applied to the diode will be

$V_{o}^{'} = 2 \times V_{o} = 2 \times 28.28V$

$\therefore$dc voltage $= \frac{2V_{o}^{'}}{\pi} = \frac{2 \times 2 \times 28.28}{22/7} = 36V$