When the frequency of the ac voltage applied to a series lcr circuit is gradually increased from a low value?
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Dear user,
It is fairly simple to understand what happens when we apply a signal to a resistance. This is because resistanceobeys Ohm's Law. However, the behaviour of Capacitors and Inductors is rather more complicated. Fortunately, we can use complex numbers and the concept of Reactance to help us understand and analyse what happens when we apply signals to circuits which contain capacitors and/or resistors.
Reactance of a Capacitor.
Consider first what happens when we apply sinewave voltage to a capacitor.

We can represent the voltage applied to the capacitor as a complex quantity

although only the real part of this is 'visible', of course.
The current flowing 'through' the capacitor (i.e. the rate at which we have to remove charge from one plate and put it onto the other) is proportional to how quickly we are changing the voltage at any instant. We can (as in the 'First 11' section of this guide) therefore use a form of differential calculus to analyse what is happening. Here, instead, we'll just say that the Capacitor's behaviour can be defined in terms of a Reactance, XC, whose value is

Where C is the capacitance value (in Farads) and f is the frequency of the wave we're applying.
To work out the current we can now use this in exactly the same way as we'd use the resistance of a resistor. Hence we can say that the current will be given by

Looking at this we can see that the 'j' which made it 'imaginary' has meant that the real (i.e. 'visible') part of the current is the Sin function, when the real part of the applied voltage is the Cos function. Hence we have a current which is out of step (by 90 degrees) with the applied voltage. This is the result we'd expect from using differential calculus since we want the peaks of the current waveform to occur when the voltage is changing most swiftly (when it is passing though zero). Note also that the magnitude of the current depends on the signal frequency. This is what we'd expect as a higher frequency means the voltage will have to change more rapidly.
Reactance of an Inductor
We can define the behaviour of an inductor in a similar way. In this case it is the rate of change of the current which is proportional to the applied voltage. So when we apply a voltage VL which is exactly the same as the VC we used earlier we get the behaviour shown in the diagram below.

To describe this behaviour can say that the reactance of the inductor, XL, will be

where L is the inductance value in Henries.
As a result we get a current of

Comparing this result with the capacitor we can see that in both cases an input cosine voltage waveform produces a sinewave current variation, but the signsof the currents produced are different. Because of this the current through the capacitor is said to lead the applied voltage, and that through the inductor is said to lag the applied voltage. We can see why these terms are used by looking at the waveforms in the above diagrams which show the sorts of patterns we'd see using an oscilloscope. The difference arises because the capacitor reactance contains a '1/j' whereas the inductor reactance has a 'j'. As a result, when we divide the voltage by the reactance this means we get a 'jj = -1' for the sine term when using a capacitor and a 'j/j = 1' for the sine term when using an inductor. The terms, 'lead' and 'lag' are used generally to indicate the 'sign' of the reactance of a circuit (i.e. capacitive or inductive)
Hope it is helpful..
It is fairly simple to understand what happens when we apply a signal to a resistance. This is because resistanceobeys Ohm's Law. However, the behaviour of Capacitors and Inductors is rather more complicated. Fortunately, we can use complex numbers and the concept of Reactance to help us understand and analyse what happens when we apply signals to circuits which contain capacitors and/or resistors.
Reactance of a Capacitor.
Consider first what happens when we apply sinewave voltage to a capacitor.

We can represent the voltage applied to the capacitor as a complex quantity

although only the real part of this is 'visible', of course.
The current flowing 'through' the capacitor (i.e. the rate at which we have to remove charge from one plate and put it onto the other) is proportional to how quickly we are changing the voltage at any instant. We can (as in the 'First 11' section of this guide) therefore use a form of differential calculus to analyse what is happening. Here, instead, we'll just say that the Capacitor's behaviour can be defined in terms of a Reactance, XC, whose value is

Where C is the capacitance value (in Farads) and f is the frequency of the wave we're applying.
To work out the current we can now use this in exactly the same way as we'd use the resistance of a resistor. Hence we can say that the current will be given by

Looking at this we can see that the 'j' which made it 'imaginary' has meant that the real (i.e. 'visible') part of the current is the Sin function, when the real part of the applied voltage is the Cos function. Hence we have a current which is out of step (by 90 degrees) with the applied voltage. This is the result we'd expect from using differential calculus since we want the peaks of the current waveform to occur when the voltage is changing most swiftly (when it is passing though zero). Note also that the magnitude of the current depends on the signal frequency. This is what we'd expect as a higher frequency means the voltage will have to change more rapidly.
Reactance of an Inductor
We can define the behaviour of an inductor in a similar way. In this case it is the rate of change of the current which is proportional to the applied voltage. So when we apply a voltage VL which is exactly the same as the VC we used earlier we get the behaviour shown in the diagram below.

To describe this behaviour can say that the reactance of the inductor, XL, will be

where L is the inductance value in Henries.
As a result we get a current of

Comparing this result with the capacitor we can see that in both cases an input cosine voltage waveform produces a sinewave current variation, but the signsof the currents produced are different. Because of this the current through the capacitor is said to lead the applied voltage, and that through the inductor is said to lag the applied voltage. We can see why these terms are used by looking at the waveforms in the above diagrams which show the sorts of patterns we'd see using an oscilloscope. The difference arises because the capacitor reactance contains a '1/j' whereas the inductor reactance has a 'j'. As a result, when we divide the voltage by the reactance this means we get a 'jj = -1' for the sine term when using a capacitor and a 'j/j = 1' for the sine term when using an inductor. The terms, 'lead' and 'lag' are used generally to indicate the 'sign' of the reactance of a circuit (i.e. capacitive or inductive)
Hope it is helpful..
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