An RLC circuit such as that of Fig. 31-7 has , , , and . (a) At what angular frequency will the current amplitude have its maximum value, as in the resonance curves of Fig. 31-16? (b) What is this maximum value? At what (c) lower angular frequency and (d) higher angular frequency will the current amplitude be half this maximum value? (e) For the resonance curve for this circuit, what is the fractional half-width ?
When driving angular frequency is equal to the natural angular frequency of the circuit, then the current amplitude is maximum because of Z=R. This condition is satisfied only when the capacitive reactance is exactly matched with inductive reactance, and it is called the condition of resonance. So, by substituting the given values in the corresponding formulae, we can find the required quantities.
The capacitive reactance of the capacitor,
The inductive reactance of the inductor,
The impedance of the LCR circuit for the driving frequency,
The current equation using Ohm’s law,
Here, R is the resistance of the resistor, C is the capacitance of the capacitor, L is the inductance of the inductor and is the driving angular frequency.
For a given resistance R and emf , I is maximum when .role="math" localid="1663000356720" .
This implies , thus using equations (1) and (2) we get the angular frequency corresponding to maximum current amplitude as follows:
For the given values, we have
Hence, the angular frequency is .
For the condition of resonance, we have ,
So the maximum value of amplitude of current is given using equation (4) as follows:
Hence, the value of the current is .
To find lower angular frequency corresponding to half of the maximum current amplitude we have the following equation using equation (3) in equation (4) as follows:
Taking the square root on both sides of the above equation, we get that
On simplifying the above equation further, we get a quadratic equation in as follows:
The two roots of this equation will give lower angular frequency and higher angular frequency corresponding to half of the maximum current amplitude.
On solving the quadratic equation, we get,
As negative frequency is not possible, so we can neglect negative sign in the second term. Thus, we get the angular frequency equation as:
Now considering the positive sign, i.e. , we can get the lower angular frequency from the above equation as follows:
We have, the above values as:
Now, substituting the above values, we can get the lower angular frequency as follows:
Hence, the value of the frequency is .
Similarly, considering the negative sign, i.e. , from equation, we can get the higher angular frequency for the above values as follows:
Hence, the value of the frequency is .
Resonance curve for this circuit will have the following fractional width as follows:
Hence, the value of the fractional width is .
Figure 31-32 shows a driven RLC circuit that contains two identical capacitors and two switches. The emf amplitude is set at , and the driving frequency is set at . With both switches open, the current leads the emf by . With switch closed and switch still open, the emf leads the current by . With both switches closed, the current amplitude is . What are (a) R, (b) C , and (c) L?
A single loop consists of inductors , capacitors , and resistors connected in series as shown, for example, in Figure-a. Show that regardless of the sequence of these circuit elements in the loop, the behavior of this circuit is identical to that of the simple LC circuit shown in Figure-b. (Hint: Consider the loop rule and see problem) Problem:- Inductors in series. Two inductors L1 and L2 are connected in series and are separated by a large distance so that the magnetic field of one cannot affect the other.(a)Show that regardless of the sequence of these circuit elements in the loop, the behavior of this circuit is identical to that of the simple LC circuit shown in above figure (b). (Hint: Consider the loop rule)
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