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## NEET & AIIMS 12th PCB Physics Demo Videos

Hello, students, welcome to the lecture video of the chapter Alternating Current. Current which varies with respect to the time over the range of values. So, in this module we will be solving the problems from a very simple concept, phase shift and time period. So, let’s get started with the first problem.

Two sinusoidal voltages of the same frequency are shown in the diagram. What is the frequency and the phase relationship between the voltage? Frequency in Hertz and phase lead of N over M in radians. So, look at these options and the right answer is b. So, carefully observe the wave forms, we have two wave forms one is M and another N. By taking these two we should find frequency. Finding the frequency is quite easy, as they have mentioned the time period if you carefully observe the wave forms. So, now let us see the solution for this problem. So, phase difference the state of vibration difference. So, when you take a wave form, so at what state it is vibrating that is what we call, its phase. When you compare the same thing with another wave form, the difference what you get, that is what we call phase difference. So, in this case we have wave forms as you see in this case. So, first wave form M, so let me mark the phase for this wave, that is 0 pi by 2 pi 3 pi by 2, 2 pi. And now let us take another wave form that is N. So, it is starting from the phase pi by 2 then what is the difference you can observe. Yes, it is minus pi by 2. It is clear phase lead of N over M is minus pi by 2. Since time period taken to complete one cycle is 0.4 sec. So, now we have got the time period. How to get the frequency then? So, getting the frequency is quite easy out of the time period, frequency is equal to 1 by time period that is going to be 1 divided by 0.4. So, the answer is going to be 2.5 Hertz. So, all you have to observe the starting point between the wave points to know about their phase shift.

So, let’s move on to the next problem. So, in case if the same problem was asked for the case of time difference then the formula is going to be time period divided by 2 pi into phase difference. This is for time difference. When you know the phase difference then only you can go for the time difference. So, let’s move onto the next question.

The instantaneous values of current and voltage in an AC circuit are respectively I is equal to 4 sin omega t and E is equal to 100 cos omega t plus pi by 3. The phase difference between voltage and current is, so, take a look at these options. So, the right answer is c. So, now in this case we have got two waves, so, let us know about the instantaneous values of current and voltage. So, instantaneous current that is i is equal to i0, i0 means peak value sin omega t. Instantaneous voltage v, v is equal to v0, v0 is the peak value of the voltage omega t plus pi. So, what is pi? Yes, it is phase difference between the current and voltage in this case. So, now do we have the same forms even in the problem? So, let us take a look at the given equations, current is given in terms of sin but the voltage are in the case EMF, so is given as the function of cos. So, now let us bring this EMF equation to the sin version. So, as you can do sin of pi by 2 plus omega t plus pi by 3. To convert the cos into sin. When you simplify, the simplified version will be 100 sin omega t plus 5 pi by 6. So, when you carefully observe these two equations. So, by knowing the instantaneous equations, what you find in the place of phase difference. It is 5 pi by 6. So, by just comparing the equations make sure both of them are in the same functions. That means sin, sin, so, in that case what do you get in the place of pi. That is your phase difference.

So, let us move onto the next problem. If the frequency of an alternating current is 50 Hertz then the time taken for the change from zero to positive peak value and positive peak value to negative peak value of current are respectively. So, take a look at these options and the right answer is a. So, we have given the frequency, by using the frequency we should find out the time taken to raise from zero to peak and positive peak to negative peak. So, to understand that let us see the wave form for this. As you have the complete wave here. So, the time taken here to complete one cycle time period. The same time period can be observed in other portion also as you can see. In this case we need to find out for zero to peak value that is, it is going to be from the origin to the peak point then the time period is going to be T divided by 4. So, now the time taken to reach zero to peak value t is equal to T divided by 4, where t is equal to 1 by f that is going to be 1 by 4f when you substitute the frequency the answer is going to be 1 by 200 sec. Is it the end of problem, no? So, you should find out one more time. So, now to understand the concept from the positive to negative peak, take a look at this wave form. In this case we have the time period in this portion and we need to find out the time taken from the positive peak to negative peak. Now, the time taken is going to be T divided by 2. So, now t dash is equal to T divided by 2. T is equal to 1 by f, that is going to 1 by 2, when you substitute the value of frequency, it is going to be 1 by 100 second.

Thank you.

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2017-04-18T04:55:14+00:00 Categories: XII-NEET & AIIMS|Tags: , , |0 Comments
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