the voltage, v, in volts, applied to an electric circuit can be modelled by the equation v=167sin(120pi)(t), where t is the time, in seconds. A component in the circuit can safely withstand a voltage of more than 120 V for 0.01 sec or less.
a) determine the length of the time that the voltage is greater than 120 V on each half-cycle.
b)is it safe to use this component in this circuit? justify your answer.
can you please explain to me clearly how I to do this q
since your amplitude is 167, that becomes your maximum voltage
let's solve when the voltage is 120
120 = 167sin(120pi)t
.71856 = sin(120(pi)t)
so 120pi(t) = .80173 (in first quadrant)
or 120pi(t) = pi - .80172 = 2.3399 (2nd quadrant)
in I:
t = .80173/(120pi) = .00212 seconds
in II:
t = .00621
so the voltage is above 120 volts from
a time of .00621-.00212 or .00441 seconds
so what do you think?
yes it can because 0.004 is less than 0.01???
yes
ok, the back of the book is wrong then because the answer says: "no" ...
and then i got confused, k thank you!
go through my arithmetic slowly, this old brain tends to make mistakes lately.
make sure your calculator is set to radians
Sure! Let's break down the problem step by step:
a) To determine the length of time that the voltage is greater than 120V on each half-cycle, we need to find the time period when the equation v = 167sin(120πt) exceeds 120V.
The equation provided is in the form v = A * sin(Bt), where A is the amplitude of the wave and B is the angular frequency. Comparing it to the given equation, we can see that the amplitude is 167V and the angular frequency is 120π.
To find the length of time that the voltage exceeds 120V, we need to find the time 't' when 167sin(120πt) > 120. Let's solve this inequality:
167sin(120πt) > 120
sin(120πt) > 120/167
To solve this, we need to find the inverse sine of both sides while keeping in mind the restrictions of the inverse sine function. Since the amplitude is 167V, the maximum value of sin(120πt) would be 167.
So, we have:
120πt > arcsin(120/167)
t > (arcsin(120/167))/(120π)
This represents the time when the voltage exceeds 120V on one half cycle. To get the complete time for both half-cycles, we need to multiply this result by 2, since a full cycle consists of two half-cycles:
Time for one half-cycle = 2 * [(arcsin(120/167))/(120π)]
b) Now let's determine if it is safe to use the component in this circuit.
The question mentions that the component can safely withstand a voltage higher than 120V for 0.01 second or less. So if the time for one half-cycle (determined in part a) is less than or equal to 0.01 second, then it is safe to use the component.
Compare the calculated time for one half-cycle with 0.01 second. If the time is less than or equal to 0.01 second, the component can be used safely. If it is greater, then it is not safe.
I hope this explanation helps!
Certainly! Let's break down the problem step by step:
a) To determine the length of time that the voltage is greater than 120V on each half-cycle, we need to find the time interval during which v exceeds 120V.
First, let's consider one half-cycle of the waveform. The equation given is v = 167sin(120πt). Since the sine function oscillates between -1 and 1, multiplying it by 167 scales it to the desired voltage range. The time period of one full cycle (or two half-cycles) for a sine function is given by T = 2π/ω, where ω is the angular frequency. In this case, angular frequency ω = 120π.
Now, we need to find the time when v exceeds 120V in each half-cycle. This occurs when the absolute value of sin(120πt) is greater than 120/167. Solving the equation |sin(120πt)| = 120/167, we get sin(120πt) = 120/167. Taking the inverse sine, we find 120πt = sin^(-1)(120/167).
Now, divide the obtained value by 120π to find the time t. This will give the time within one half-cycle when v exceeds 120V. Multiply this value by 2 to get the total duration in both half-cycles.
b) To determine whether it is safe to use this component in the circuit, we need to compare the calculated duration from part a) with the safe limit provided, which is 0.01 sec.
If the calculated duration is greater than 0.01 sec, then it is not safe to use this component. However, if the calculated duration is less than or equal to 0.01 sec, then it is safe to use this component.
I hope this explanation helps you in solving the problem. Let me know if you have any further questions!