Check Your Understanding If a European ac voltage source is considered, what is the time difference between the zero crossings on an ac voltage-versus-time graph?
Check Your Understanding If a European ac voltage source is considered, what is the time difference between the zero crossings on an ac voltage-versus-time graph?
Check Your Understanding If a European ac voltage source is considered, what is the time difference between the zero crossings on an ac voltage-versus-time graph?
Expert Solution & Answer
To determine
The time difference between the zero crossings of an ac voltage versus time graph for an European voltage source.
Answer to Problem 15.1CYU
The time between two zero crossing is 8.34×10−3s.
Explanation of Solution
Formula Used:
The relation between the time period and frequency is,
T=1f
The expression for the time period between two zero crossing is.
t=T2
Calculation:
The European frequency is 60Hz.
The value of the time period of an European voltage source is calculated as,\
T=160 Hz=0.017s
The value of the time between two zero crossing is calculated as,
t=0.017s2=8.34×10−3s
Conclusion:
The time between two zero crossing of an European voltage source is 8.34×10−3s.
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An AC Power Supply Produces a
maximum Voltage Symax = 94V : This
Power Supply is connected to a resistor
R = 20-852- and the current and resistor
Voltage are measured with an Ideal AC
ammeter and Voltmeter. An ideal ammeter
has Zero resistance, and an Ideal Poltmeter
has Infinite resistance.
a) What is the reading
on
the Ammeter ?
The maximum voltage in AC circuit is 100V . What is the effective voltage in circuit?
Example:
(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak +V,
power consumption rate of a 60.0-W AC light bulb?
For (a):
-Vo
Solving the equation Vrms= for the peak voltage Vo and substituting the known
value for Vms gives:
Vo=2-vZVrms=1.414(120 V)= 170 V.
This means that the AC voltage swings from 170
V to -170 V and back 60 times every second. An
equivalent DC voltage is a constant 120 V.
For (b):
Peak power is peak current times peak voltage. Thus,
Figure 2. The potential difference V
between the terminals of an AC voltage
So the power swings from zero to 120 W on
hundred twenty times per second (twice each
cycle), and the power averages 60 W.
fluctuates
shown.
The
Po= loVo = 2(; loVo) = 2Pave.
source
as
mathematical expression for V is given by V
= Vosin2nft
Po = 2(60.0 W) = 120 W.
Average
power
Activity:
Given a transmission line having a line resistance of 5.000 sove for the following.
Use the rubrics as your guide in presenting your solution…
Essential University Physics: Volume 1 (3rd Edition)
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