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In Figure, R=14.0Ω,C=6.20μ¹ó,and L=54.0mH, and the ideal battery has emf ε=34.0V. The switch is kept at a for a long time and then thrown to position b.

(a)What is the frequency? (b)What is the current amplitude of the resulting oscillations?

Short Answer

Expert verified
  1. The frequency is 275Hz.
  2. The current amplitude of the resulting oscillation is 0.365A.

Step by step solution

01

The given data

  1. Resistance attached in the circuit R=14Ω,
  2. Inductance of the inductor, role="math" localid="1663075735599" L=54.0mHor54×10-3H
  3. Capacitance of a capacitor,C=6.20μ¹óor6.20×10-6F
  4. The emf voltage, ε=34.0V
02

Understanding the concept of the mechanism of the LC circuit

When a switch is kept at a point the capacitor is charged. After that, the switch is thrown to position b; then the circuit is an LC circuit. From that, we can calculate the angular frequency and frequency. Now capacitor is charged up to maximum voltage, and we can calculate the maximum charge on the capacitor. From that, we can calculate the current amplitude by taking the relation between the current and charge and angular frequency.

Formulae:

The angular frequency of an LC oscillation, Ó¬=1LC(i)

The charge of the capacitor, Q=CV(ii)

The current amplitude of the LC circuit, i=Ó¬Q(iii)

The frequency of an oscillation,f=Ó¬2Ï€(iv)

03

a) Calculation of the frequency

When the switch is thrown to position b, the circuit is an LC circuit. The frequency of the oscillation can be given using equation (i) in equation (iv) as follows:

f=12πLC=12π54×10-36.20×10-6=275Hz

Hence, the value of the frequency is 275Hz.

04

b) Calculation of the current amplitude of the resulting oscillation

After that, the switch is thrown to the point b so that the capacitor is chargedV=34.0V.Now, the maximum charge on the capacitor can be calculated using equation (ii) as follows:

Q=6.20×10-6×34.0=2.11×10-4C

So now, the current amplitude can be calculated using equation (iv) in equation (iii) as follows:

i=2πfQ=2π×275×2.11×10-4=0.365A

Hence, the value of the current amplitude is 0.365A.

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Most popular questions from this chapter

An RLC circuit such as that of Fig. 31-7 hasR=5.0Ω,C=20.0μ¹ó,L=1.0H, andεm=30.0V. (a) At what angular frequencyÓ¬dwill 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 frequencyÓ¬d1and (d) higher angular frequencyÓ¬d2will the current amplitude be half this maximum value? (e) For the resonance curve for this circuit, what is the fractional half-width(Ó¬d2-Ó¬d1)/Ó¬?

An alternating source with a variable frequency, an inductor with inductance L, and a resistor with resistance Rare connected in series. Figure gives the impedance Zof the circuit versus the driving angular frequency Ó¬d, with the horizontal axis scale set by (Ó¬d)s=1600rad/s. The figure also gives the reactance XLfor the inductor versus Ó¬d. (a) What isR? (b) What isL?

In an oscillating LCcircuit L=1.10mH, andC=4.00μF. The maximum charge on the capacitor is3.00μC. Find the maximum current.

In Fig. 31-7, R=15.0Ω, C=4.70μF, and L=25.0mH. The generator provides an emf with rms voltageVRCL=75.0Vand frequency f=550Hz. (a) What is the rms current? What is the rms voltage across (b) R, (c) C, (d) L, (e) C and L together, and (f) R, C, and L together? At what average rate is energy dissipated by (g) R , (h) C, and (i) L?

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