The current is defined as the quantity of charge Q that passes through a certain location in a time 
:
Using the data of the problem, we find:
Using the data of the problem, we find:
A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input
1 answer:
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 - ( )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
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Answer:
The magnitude of the static frictional force is 1200 N
Explanation:
given information :
radius, r = 0.380 m
applied-torque, τ1 = 456 N
The car has a constant velocity, thus the acceleration is zero
α = 0
Στ = I α
τ1 - τ2 = I α
τ2 = counter-torque
τ1 - τ2 = 0
τ1 = τ2
r x = τ1
= the static frictional force (N)
= τ1 /r
= 456 N/0.380 m
= 1200 N