Respuesta :
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - ( [tex]\frac{1}{2fRC}[/tex])V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V
Ripple factor is the RMS value of an alternating current component divided by the rectified output's average value. The peak-to-peak ripple voltage will be 2V.
what is the ripple factor?
It is defined as the RMS value of an alternating current component in the rectified output is divided by the rectified output's average value. It is a dimensionless quantity with a value less than unity at all times.
The given data in the problem is;
Voltage of full-wave rectifier = 120 V
input produces = 60 Hz
peak of voltage= 15 V
load value = 1.0 kV
DC output voltage = 14 V
[tex]\rm V_R[/tex] is the peak-to-peak ripple voltage = ?
The output voltage is given by ;
[tex]\rm V_0= 1-\frac{1}{(2fRC)V_1} \\\\ \frac{V_0}{V_1}= 1-\frac{1}{(2fRC)}\\\\ \frac{1}{(2fRC)}= 1-\frac{V_0}{V_1}[/tex]
The value of capacitance is found by;
[tex]\rm C= 2 \times 240 \times 1 ((1-\frac{14}{15}))^{-1 } \\\\ \rm C=62.2 \mu f[/tex]
The peak to peak ripple voltage is found by;
[tex]\rm V_R = (\frac{1}{fRc} )V_1 \\\\ V_R = (\frac{1}{12. \times 1 \times 62.2} )15 \\\\ \rm V_R =2 V[/tex]
Hence the peak-to-peak ripple voltage will be 2V.
To learn more about the peak to peak ripple voltage refer to;
https://brainly.com/question/14789569
