In the figure, the battery has an emf of 15.0 V, the inductance is 2.98 mH, and

Question

In the figure, the battery has an emf of 15.0 V, the inductance is 2.98 mH, and the capacitance is 8.50 pF. The switch has been set to position a for a long time so that the capacitor is charged. The switch is then thrown to position b, removing the battery from the circuit and connecting the capacitor directly across the inductor.

(A) Find the frequency of oscillation of the circuit.

Conceptualize When the switch is thrown to position b, the active part of the circuit is the right-hand loop, which is an LC circuit.

Categorize We use equations developed in this section, so we categorize this example as a substitution problem.

Use ? = 1 / ?LC to find the frequency:

Substitute numerical values:

(B) What are the maximum values of charge on the capacitor and current in the circuit?

Analyze

Find the initial charge on the capacitor, which equals the maximum charge:

Use ??Qmax sin(?t) = ?Imax sin(?t) to find the maximum current from the maximum charge:

Imax = ?Qmax = 2?f Qmax

I Cannot Solve the final Question. Please Help.

First the capacitor is fully charged with the switch set to position a. Then, the switch is thrown to position b and the battery is no longer in the circuit.

(A) Find the frequency of oscillation of the circuit.

Conceptualize When the switch is thrown to position b, the active part of the circuit is the right-hand loop, which is an LC circuit.

Categorize We use equations developed in this section, so we categorize this example as a substitution problem.

Use ? = 1 / ?LC to find the frequency:

f = ? 2? = 1 2? ?LC

Substitute numerical values:

f = 1 2?[(2.98 10-3 H)(8.50 10-12 F)]1/2

f =   Hz

(B) What are the maximum values of charge on the capacitor and current in the circuit?

Analyze

Find the initial charge on the capacitor, which equals the maximum charge:

Qmax = C ?V = (8.50 10-12 F)(15.0 V)

Qmax =   C

Use ??Qmax sin(?t) = ?Imax sin(?t) to find the maximum current from the maximum charge:

Imax = ?Qmax = 2?f Qmax

Imax = (2? ? 106 s-1)(2.98 10-10 C)

Imax =  
Your response differs from the correct answer by more than 10%. Double check your calculations.

What if the charge on the capacitor is found to be 62% of the maximum value at a particular instant, what is the current (magnitude only) flowing through the circuit at that instant? Give your answer as a percentage to the maximum current value.

x =   %
How can we use the concept of energy conservation to solve this problem? Your response differs from the correct answer by more than 10%. Double check your calculations.

I Cannot Solve the final Question. Please Help.

10 -000

Explanation / Answer

I assume that you have already solved first 3 parts ( frequency, Qmax , Imax )

So, now, Q' =  62% of Qmax = 0.62*1.275e-10

  Q' = 7.905e-11

Now use this value to calculate Imax ,

Imax' = 2*pi*f*Q'  = 2*pi*106*7.905e-11

Imax'   = 4.97e-4 amps

Now, we need to get this into percent

Imax' = x% of Imax

4.97e-4 = x/100*8.011e-4

Imax' = 62% of Imax

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