What if the charge on the capacitor is found to be 68% of the maximum value at a

Question

         

    

    

    

                           What if the charge on the capacitor is found to be 68% 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 =      %

f =                                  ? 2?                              =                                  1 2?  ?LC                              In the figure, the battery has an emf of 10.0 V, the inductance is 3.00 mH, and the capacitance is 8.60 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.

Explanation / Answer

For the last question: Use the conservation of energy.

U=UE + UB = q2/(2C) + (1/2)*L*i2 = (Qmax2)/(2C)

You have already calculated Qmax, and have it equal to 8.6*10-11 Coulombs.

L= 3*10-3 H and C = 8.6*10-12 given by the problem.

q=.68*Qmax = .68 * 8.6*10-11 = 5.85*10-11 C

Therefore from the equation above:

(5.85*10-11)2 / (2*8.6*10-12) + (1/2) * (3*10-3) * i2 = (8.6*10-11)2 / (2*8.6*10-12)

Solving for i in this case:

i (with q at 68% of the maximum charge) = 3.92*10-4 Amperes.

Imax was calculated to be .00068 A in the earlier problem.

Therefore:

i/Imax = 3.92*10-4 / .00068 = 0.5771

So, the current at the same time at which the charge is at 68% of maximum charge is 57.71% of the maximum current.

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