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3 years ago

6

Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and

a 1k ohm load are connected the DC output voltage is 14V. What is... The value of the capacitor? The value of the peak to peak ripple voltage?

1 answer:

Alborosie3 years ago

5 0

Answer:

$V_{pp}=2V$

Explanation:

Source Voltage $V= 120V$

Frequency $f=60Hz$

Peak output voltage $Vp=15V$

Peak Output Voltage with filter $V_p'=14V$

Generally the equation for Peak to peak voltage is mathematically given by

$V_p'=V_p-\frac{V_{pp}}{2}$

Therefore

$V_{pp}=2(V_p-v_p')$

$V_{pp}=2(15-14)$

$V_{pp}=2V$

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An aluminum alloy [E = 73 GPa] cylinder (2) is held snugly between a rigid plate and a foundation by two steel bolts (1) [E = 18

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$\sigma_A = 58.43 N/mm^2$

Explanation:

Given data:

length of Steel bolt $L_1 = 335 mm$

Length of aluminium cylinder $L_2 = 275 mm$

Pitch of bolt p = 1mm

Modulus of elasticity of steel E = 215 GPa

Modulus of elasticity of aluminium = 74 GP

Area of bolt $= \frac{\pi}{4} 14^2 = 153.93 mm^2$

Area of cylinder = 2300 mm^2

n =1

By equilibrium

$\sum F_y = 0$

$P_A -2P_S = 0$

$P_A =2P_S$

By the compatibility

$\delta _s + \delta_A = nP$

Displacement in steel is $\delta_s = \frac{P_sL_s}{E_sA_s}$

Displacement in Aluminium is $\delta_A = \frac{P_AL_A}{E_AA_A}$

from compatibility equation we have

$\frac{P_sL_s}{E_sA_s} + \frac{P_AL_A}{E_AA_A} = nP$

$\frac{P_s\times 335}{180\times 10^3\times 153.93} + \frac{P_A\times 275}{73\times 10^3\times 2300} = 1\times 1$

$1.20\times 10^[-5} P_s + 1.44\times 10^{-6}P_A = 1$

substitute $P_A =2P_S$

$1.20\times 10^[-5} P_s + 1.44\times 10^{-6} (2\times P_s) = 1$

$1.488\times 10^{-5} P_s = 1$

$P_s = 67204.30 N$

$P_A = 134,408.60 N$

Stress in Aluminium $\sigma = \frac{P_A}{A_A}$

$= \frac{134,408.60}{2300}$

$\sigma_A = 58.43 N/mm^2$

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3 years ago