Question Video: Determining the Frequency of the Output Potential Difference of a Generator | Nagwa Question Video: Determining the Frequency of the Output Potential Difference of a Generator | Nagwa

# Question Video: Determining the Frequency of the Output Potential Difference of a Generator Physics • Third Year of Secondary School

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The graph shows the output potential difference of a generator over time. What is the frequency of the output potential difference?

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### Video Transcript

The graph below shows the output potential difference of a generator over time. What is the frequency of the output potential difference?

To answer this question, we need to use the graph we’ve been given to work out the frequency of the output potential difference of a generator. We can see that this graph has the output potential difference, in units of volts, plotted on the 𝑦-axis and the time, in units of milliseconds, plotted on the 𝑥-axis.

Recall that the frequency of an oscillation can be calculated using the formula frequency equals one divided by the period of oscillation. This is often written as 𝑓 equals one over capital 𝑇, where 𝑓 is the frequency and capital 𝑇 is the period. So, before we can find the frequency of the output potential difference, we first need to find the period.

The period is simply the time taken for one complete cycle of the oscillation to occur. We can use the graph to work this out. We could look at any oscillation cycle shown on this graph, but it’s probably easiest to look at this first one here. Notice that we’re including this starting point here, this maximum value, this minimum value, and this zero point here in one single oscillation. It’s important that we include the whole cycle in our working. Otherwise, we wouldn’t calculate the right value for the period.

The period, or the time taken for this oscillation to occur, is simply equal to the time difference between this point here on the time axis and this point here. So all we really need to do is read these values from the graph. The scale marked on the horizontal axis tells us that 10 of these vertical grid lines corresponds to 100 milliseconds. So we know that each individual line must represent 10 milliseconds.

The oscillation begins at a time of zero milliseconds, and it ends here at the second grid line. Since one vertical grid line corresponds to 10 milliseconds, then the end of this first cycle after two grid lines must correspond to 20 milliseconds. The period of oscillation is equal to the time difference between these two times, and so the period is equal to 20 milliseconds minus zero milliseconds. This is simply equal to 20 milliseconds.

However, before we can use this value to calculate the frequency of the oscillation, we first have to convert it into units of seconds. Recall that the unit prefix milli- is equivalent to a factor of 10 to the negative three. So 20 milliseconds is equal to 20 times 10 to the negative three seconds, which we can write more simply as 𝑇 equals 0.02 seconds. So the oscillation period of the output potential difference is 0.02 seconds.

Let’s clear some space on screen so we can substitute this into our equation for frequency. Substituting our value for the period into this equation, we have that the frequency of the oscillation is equal to one divided by 0.02 seconds.

Before we calculate this number, let’s take a moment to think about the units of the quantities. We have no units in the numerator of this expression, and in the denominator we have units of seconds. Overall, this gives us units of inverse seconds or seconds to the negative one.

However, frequency is more commonly expressed in units of hertz. One hertz is equal to one inverse second. So we can just replace the units of inverse seconds in our expression by units of hertz. Now, all that’s left to do is to calculate the value of one over 0.02 hertz. This gives us our final answer of 50 hertz. We have found then that the frequency of the output potential difference of the generator is equal to 50 hertz.

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