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Fundamentals Of Physics
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Short Answer

An RLC circuit such as that of Fig. 31-7 has R=5.0 Ω, C=20.0 μF, L=1.0 H, and εm=30.0 V. (a) At what angular frequency ωd 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 ωd1 and (d) higher angular frequency ωd2 will the current amplitude be half this maximum value? (e) For the resonance curve for this circuit, what is the fractional half-width (ωd2-ωd1)/ω ?

  1. Angular frequency ωd corresponding to maximum current amplitude is 224 rad/s.
  2. Maximum value of amplitude of current is 6.0 A.
  3. Lower angular frequency ωd1 corresponding to half of the maximum current amplitude is 219 rad/s.
  4. Higher angular frequency ωd2 corresponding to half of the maximum current amplitude is 228 rad/s.
  5. Fractional half width is 0.040.
See the step by step solution

Step by Step Solution

Step 1: The given data

  1. Amplitude of emf, εm=30.0 V
  2. Resistance, R=5.0 Ω
  3. Capacitance, C=20.0 μF
  4. Inductance, L=1.0 H

Step 2: Understanding the concept of frequency of LC circuit

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,

XC=1ωdC ...(1)

The inductive reactance of the inductor,

XL=ωdL ...(2)

The impedance of the LCR circuit for the driving frequency,

Z=R2+XL-Xc2 ...(3)

The current equation using Ohm’s law,

I=εmZ ...(4)

Here, R is the resistance of the resistor, C is the capacitance of the capacitor, L is the inductance of the inductor and ωd is the driving angular frequency.

Step 3: a) Calculation of the angular frequency

For a given resistance R and emf εm, I is maximum when .role="math" localid="1663000356720" χL-χC=0.

This implies XL=XC, thus using equations (1) and (2) we get the angular frequency ωd corresponding to maximum current amplitude as follows:

ωdL=1ωdC ωd2=1LC =1LC

For the given values, we have

ωd=11.0 H×20.0×10-6 F =10320.0 rad/s =223.6 rad/s 224 rad/s

Hence, the angular frequency is 224 rad/s.

Step 4: b) Calculation of the maximum current

For the condition of resonance, we have Z = R,

So the maximum value of amplitude of current is given using equation (4) as follows:

I=30.0 V5.0 Ω =6.0 A

Hence, the value of the current is 6.0 A.

Step 5: c) Calculation of the lower angular frequency

To find lower angular frequency ωd1 corresponding to half of the maximum current amplitude we have the following equation using equation (3) in equation (4) as follows:

I2=εmR2+XL-XC2 εm2R=εmR2+XL-XC2 12R=1R2+XL-XC214R2=1R2+XL-XC2

Solving further

4R2=R2+XL-XC23R2=XL-XC2

Taking the square root on both sides of the above equation, we get that

localid="1663002854238" ±3·R=XL-XC =ωdL-1ωdC (from equations (1) and (2))

On simplifying the above equation further, we get a quadratic equation in ωd as follows:

LCωd2±3·RCωd-1=0

The two roots of this equation will give lower angular frequency ωd1 and higher angular frequency ωd2 corresponding to half of the maximum current amplitude.

On solving the quadratic equation, we get,

ωd=-±3·RC±3RC2+4LC2LC

As negative frequency is not possible, so we can neglect negative sign in the second term. Thus, we get the angular frequency equation as:

ωd=-±3·RC+3RC2+4LC2LC

Now considering the positive sign, i.e. +3·RC, we can get the lower angular frequency ωd1. from the above equation as follows:

ωd1=-3·RC+3RC2+4LC2LC

We have, the above values as:

RC=5.0 Ω ×20.0×10-6 F =10-4 ΩF

And

LC=1.0 H ×20.0×10-6 F =0.2×10-4 HF

Now, substituting the above values, we can get the lower angular frequency as follows:

ωd1=-3×10-4Ω F+3×10-8 ΩF2+4×0.2×10-4 HF2×0.2×10-4 HF = 219 rad/s

Hence, the value of the frequency is 219 rad/s.

Step 6: d) Calculation of the higher angular frequency

Similarly, considering the negative sign, i.e. -3·RC, from equation, we can get the higher angular frequency ωd2 for the above values as follows:

ωd2=+3·RC+3RC2+4LC2LC =3×10-4ΩF+3×10-8 ΩF2+4×0.2×10-4 HF 2×0.2×10-4HF =228 rad/s

Hence, the value of the frequency is 228 rad/s.

Step 7: e) Calculation of the fractional half-width

Resonance curve for this circuit will have the following fractional width as follows:

(ωd2-ωd1)ω=228.0 rad/s-219.0 rad/s224.0 rad/s =0.040

Hence, the value of the fractional width is 0.040.

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