A 55 µF capacitor with an initial energy of 0.9 J is...

A 55 µF capacitor with an initial energy of 0.9 J is discharged through a 5 ?? resistor. What is the initial charge on the capacitor? The charge, $Q_0$ = Units Select an answer What is the current through the resistor when the discharge starts? The current, $I_0$ = Units Select an answer Determine the potential difference across the capacitor and the rate at which the thermal energy is dissipating in the resistance 2.8 min after the discharge starts. The potential difference across C, $V_c$ = Units Select an answer The power dissipated in R, P = Units Select an answer

Transcript

00:01 Potential energy stored in a capacitor is equal to q square, where q is the charge stored in the capacitor over two times capacitors of the capacitor.

00:14 Now we are provided with the initial potential energy in the capacitor.

00:20 So to find the initial charge, we can use this equation.

00:25 So q0 will be twice c times, times excuse me the initial potential energy stored of stored in the capacitor and we need to take a root over now let's substitute the values and solving this we get q0 to be 1020 power minus three columns so that's the initial charge stored in the capacitor now in a rc circuit that is discharging with respect to time, the charge as a function of time can be given by this equation where t is the time, q is the charge in the capacitor at time d, and tau here is the capacitive rate constant that is equal to r times c.

01:44 Now we know that current is the derivative of charge with respective time.

01:49 So i will be equal to negative dq over d t and the sign negative is because q is decreasing with time.

01:59 This is a this is the equation for it is charging rc circuit therefore q is decreasing with time therefore dq over dd will lead to a negative value and in order to make that positive we need the current to be negative of the derivative.

02:17 This is equal to q0 over tau e to the power minus t over tau.

02:23 So this is positive now.

02:26 And from here we can say that the initial current that is at t equals to zero, the current will be equal to q0 over tau.

02:38 So this basically came by substituting t to be equals to zero.

02:42 And now that we have this, we can first we can then calculate the time constant that is equal to the resistance times capacitance so i'm plugging in all the values so this gives the capacitive time constant to be once again and now we can find the initial current using this equation this one so we have q0 from the first part and we calculated tau just now.

03:25 So we will substitute those two values and get the current to be 1 .0 times 10 to the power minus 3 a appears.

03:48 Now in the second part we have to find the voltage.

03:53 So we know that q is equal to q0 times e to the power minus t over tau.

04:01 And v from faradet's log can be set to be the charge stored in the capacitor over capacitance.

04:10 So now we can substitute this over here.

04:18 So this means that v will be equal to q0, e to the bar minus t over tau over c.

04:24 So i'm just substituting the q from this expression over here.

04:33 So now plugging in all the values and this gives the voltage to be equal to 1 .0 times 10 to the power 3 volts times e to the power minus minus t.

05:05 Where t is measured in seconds because our tau is in seconds.

05:11 So t has to be in seconds in order to make this ratio.

05:15 Unitless so that we can apply the exponential function to it.

05:23 So that's the expression for the voltage...

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