The parallel-plate capacitor in the circuit shown to the right is charged and th

Question

The parallel-plate capacitor in the circuit shown to the right is charged and then the switch is closed. At the instant the switch is closed, the current measured through the ammeter is I. After a time of 2.4s elapses, the current through the ammeter is measured to be 0.60I, and the switch is opened. A substance with a dielectric constant of 1.5 is then inserted between the plates of the capacitor, and the switch is once again closed and not reopened until the ammeter reads zero current.

a. Find the period of time that elapses between when the switch is closed the second time and when the ammeter reads a current

into thermal energy by the resistor.

SP 11 The parallel-plate capacitor in the circuit shown to the right is charged and then the switch is closed. At the instant the switch is closed, the current measured through the ammeter is I. After a time of 2.4s elapses, the current through the ammeter is measured to be 0.601, and the switch is opened. A substance with a dielectric constant of 15 is then inserted between the plates of the capacitor, and the switch is once again closed and not reopened until the ammeter reads zero current. a. Find the period of time that elapses between when the switch is closed the second time and when the ammeter reads a current of 0.21 b. At the end, all of the electrical potential energy is gone from the capacitor. Find the fraction of this energy that was converted into thermal energy by the resistor. 200

Explanation / Answer

For the time t = 0 to t = 2.4s,

the current in the curcuit :

i = I*e^(-t/T)

At t= 2.4s, i = 0.6*I

So, 0.6*I = I*e^(-2.4/T)

So, T = 4.7 s

Now, T = RC = 4.7

Now, when the dielectric(of dielectric constant k= 1.5 ) is inserted, Capacitance(C) changes to k*C

and the voltage of the capacitor changes to V/k <---- here V = voltage at t = 2.4s

Now, the current in the circuit after t' = t - 2.4 = 0,

i' = I'*e^(-t'/(R*C/k)) = I'*e^(-t'k/RC)

Now, just after t = 2.4s, the current in the circuit, = 0.6*I/k = 0.6*I/1.5

So, i' = (0.6*I/1.5)*e^(-t'*1.5/RC)

So, for current , i = 0.2*I ,

0.2*I = (0.6*I/1.5)*e^(-t'*1.5/RC)

So, 0.5 = e^(-t'*1.5/(4.7)) <------- RC = 4.7

So, t' = 2.17 s <-------answer

b)

At the start let the charge across the capacitor be Q

So, just before t = 2.4s, the charge was 0.6*Q

So, energy stored now = (0.6Q)^2/2C = 0.18*Q^2/C

Just after t=2.4 s (when dielectric is intorduced) , the capacitance increases to 1.5*C

So, energy stored now = (0.6Q)^2/(2*1.5*C) = 0.12*Q^2/C

So, change in energy = 0.06*Q^2/C

So, energy lost to thermal energy after a long time = Q^2/2C - 0.06*Q^2/C = 0.44*Q^2/C

So, fraction of energy converted to thermal energy by resistor = 0.44*Q^2/C / (Q^2/2C) = 88 percent <---- answer

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