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Fundamentals Of Physics
Found in: Page 936

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Short Answer

In an oscillating LC circuit, L= 3.00 mHand C=2.70 μF. At t= 0 the charge on the capacitor is zero and the current is 2.00 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time t> 0 is the rate at which energy is stored in the capacitor greatest, and (c) What is that greatest rate?

  1. The maximum charge that will appear on the capacitor is 1.8×10-4 C.
  2. The rate at which energy stored is the greatest in the capacitor is 7.07×10-5 s.
  3. The greatest rate is 66.7 W.
See the step by step solution

Step by Step Solution

Step 1: The given data

  1. The inductance L=3.00 mH or 0.003 Hand capacitanceC=2.70 μF or 2.70×10-6F are present in the oscillating LC circuit.
  2. At t = 0, the charge on the capacitor, q = 0 andi=2.00 A.

Step 2: Understanding the concept of charge and energy of LC circuit

By using the concept of charge, which is the function of time, we first find the current, which is also a function of time. Using the initial condition, we can find the maximum charge that will appear on the capacitor. Further, by using the equation of energy stored in the capacitor, we can find the time rate at which energy stored in the capacitor is the greatest.


The current equation related to the charge rate, i=dqdt (i)

The angular frequency of an LC oscillation, ω=1LC (ii)

The electrical energy stored by the capacitor, UE=q22C (iii)

The charge is a function of time, q=Qsinωt (iv)

Where Q is the maximum charge on the capacitor and ω is the angular frequency of oscillation.

The angular frequency of an oscillation, ω=2πT (v)

Step 3: a) Calculation of the maximum charge on the capacitor  

Substituting equation (iv) in equation (i), we get the current equation as follows:


At time t = 0 then,



i=ω Q

So the charge Q is given by substituting the value of equation (ii) in the above equation as follows:

Q=iLC=2.00×3.00×10-3×2.70×10-6=2.00×9×10-5=1.8×10-4 C

Hence, the value of the charge is 1.8×10-4 C.

Step 4: b) Calculation of the time at which rate of the energy stored is greatest

By substituting the value of charge from equation (iv) in equation (iii), we get the energy equation as follows:


And its rate of change is now given as:


But by using the trigonometric identity, we can write,



cosa sina=12sin2a

Thus, the energy rate equation becomes:

dUEdt=Q2ωsinωtcosωt C=Q2ωsin2ωt2C...........................(a)

The greatest rate of change occurs when sin2ωt=1, i.e.,

2ωt=π2t=π4ω=π4 LC (from equation (ii))=π43×10-32.70×10-6=π4×9×10-5=7.07×10-5s


Hence, the value of time is 7.07×10-5 s.

Step 5: c) Calculation of that greatest energy rate

From equation (a), we can get the greatest rate of energy stored for sin2ωt=1 as follows:

dUEdt=Q2ω2C=2πQ22TC (ω=2πTfrom equation (v))=πQ2TC....................(b)

Now, using equation (ii) in equation (v), the period of oscillations can be given as follows:


Now, substituting the above value in equation (b) with other given data, we can get the greatest rate value as follows:

dUEdt=π1.80×10-425.655×10-4×2.70×10-6=66.7 W

From this result, we can conclude that the energy in the capacitor is indeed increasing at t=T8.

Hence, its greatest rate is 66.7 W.

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