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JEE MAIN & ADVANCE 12th PCM Physics -Alternating Current Module-2 Demo Videos

Hello students, welcome back to the chapter of alternating current. In the previous few modules we have been discussing that how we can you phasor techniques to solve the questions where we have the combination of more than one circuit elements that is an LCR circuits. That means in a circuit we have all three, the inductor, the capacitor and the resistor. So, let’s take the discussion further and consider an AC circuit with L, C and R as shown in this particular figure. Now, let’s assume that at certain instant of time the current is drawing in the clockwise direction and the potential difference across these three elements can be written as VL, VC and VR. Now, what we already know from our previous discussions is that in case of a resistor the current and emf are in same phase, whereas in case of a capacitor the current leads the emf by pi by 2 and in case of an inductor the current is lagging behind the emf by pi by 2.

Now, we have been discussing everything till now taking the voltage of the source Epsilon as the reference. What if we change our reference to the current. So, here the noticeable thing before we move ahead is that at any particular instant the value of emf would be the sum of all three voltages. That means at an instant the voltage would be ruling by the Kirchoff’s law and not by the phasor diagram. So, let’s say that at any instant the current is given by i0 sin of omega t plus 5 and let us continue the discussion that what happens if we take the current as our reference. Now, if we take current as the reference nothing happens to the emf for a resistor because for a resistor the emf and the current are in same phase. But in case of a capacitor the current actually leads the emf and therefore we can say that the emf or the voltage actually lags behind the current and therefore in case of a capacitor we can write VC is equal to the maximum value multiplied by sin of omega t plus 5 which is same for the current minus of pi by 2 since the current leads or with respect to the current the voltage lags behind. Now, if you take inductor for example what we will get is that the current actually lags behind the voltage and therefore if the current is the reference then we can say the voltage leads ahead by an angle of pi by 2. And therefore in the angle value of the sin we will get omega t plus 5 plus of pi by 2. So, please note here we are discussing if we take the current as the reference. In all our previous discussions we have taken the voltage as the reference.

Now, let’s continue our discussions about these LS, LCR circuit. What we can note here is that the value of phasor diagrams or the technique of phasor diagram is well applicable for RMS voltage which can be given by the resultant of VC minus VL and VR. Now, to find out the maximum values we will use simply the Ohm’s law which is given by V by R. So, for resistance it is given by V is equal i0 R sin of omega t plus 5. For the value of capacitor we can say that VC is equal to i0 into Xc and the voltage actually lags behind the current by pi by 2 and in case of inductor we can write VL would be equal to i0 into XL sin of omega t plus 5 plus pi by 2. That means the voltage leads the current by an angle of pi by 2, which is same as the current lags behind the voltage at an angle of pi by 2. Now, you can write down the value of current by the resultant of the voltages and the impedance of the circuit. Let’s see how it is done. Now, we can write down here that Epsilon0 would be the resultant of these three when we talk about the RMS value and therefore Epsilon0 will be equal to under root of VR square plus VL minus VC square, which then can be written as i0 into R plus i0 into XL minus XC square. Taking i0 common we get this particular value and on simplification we can write down i0 is equal to e0 by Z, which basically is what we have discussed earlier that Ohm’s law is applicable on impedance.

Now, let’s try to find out the power developed across such a circuit. We know instantaneous power nothing but Epsilon into y. Putting down the value that we have here we will get that it is equal to Epsilon0 i0 sin of omega t multiplied by sin of omega t plus 5, where 5 can be found out using this particular diagram or the phasor methods that we know. P average whereas is given by ERMS into IRMS multiplied by cos 5, where 5 again is the angle between the current and the emf which can be found out using the phasor diagrams, which in this particular case can be written as R by Z and therefore power factor for an LCR circuit is given by R by Z.

I hope you’ve understood the module. We will take our discussion further in the next coming section, thank you.

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2017-04-18T04:56:02+00:00 Categories: IIT - JEE Main & Advanced XII - Physics||0 Comments
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