First slide

Application of p-n junction diode rectifier

Question

Let the input to a full-wave rectifier be e = 50 sin (314t) volt and its diode and load resistances are 100 Ω and 1 Ω, respectively. Consider the following statements.
(I) its pulse frequency of output voltage is 100
(II) its input power is 1136 mW
(III) its output power is 838 mW
(IV) its efficiency is 81.2%
Which of the options is correct based on the above information?

Difficult

A

(I), (III)
B

(I), (II)
C

(I), (II), (III)
D

(I), (II), (III), (IV)

Solution

Given that,
e = 50 sin (314t)
$e_{\circ} = 50 volt, ω = 314 Hz$
(1) in a full wave rectifier plus frequency
$f_{\circ} = 2 f$
$f_{\circ} = 2 \frac{ω}{2 π} = \frac{ω}{π}$
$f_{\circ} = \frac{314}{3.14}$
$f_{\circ} = 100 Hz$
(2) input power
$p_{i} = \frac{e^{2}}{2 (r + R)}$
$P_{i} = \frac{50^{2}}{2 (1 + 100)} = 1136 mW$
(3) output power
$P_{\circ} = {[\frac{2 V_{0}}{π (r_{f} + R_{L})}]}^{2} \times R_{L}$
$P_{\circ} = \frac{2 \times 2 \times 50 \times 50 \times 1}{π \times π \times 1.1 \times 1.1} = 838 mW$
(4) efficiency
$η = \frac{output power}{input power} \times 100$
$η = \frac{838}{1136} \times 100 = 73 .76 %$

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