Given capacitance & energy dissipated, find the charge Q

[courteous]

Homework Statement

A capacitance of $$2\mu F$$ having a charge $$Q_0$$ is switched into a series circuit of $$10\Omega$$ resistance. Find $$Q_0$$ if the total energy dissipated in the resistance is $$3.60 mJ$$.

Ans. $$120 \mu C$$

Homework Equations

$$Q=CV$$

The Attempt at a Solution

$$E=\int_0^\infty p(t)dt=\int_0^\infty \frac{v(t)^2}{R}dt=3.60 mJ$$

I need to extract $$V$$ and use it in the equation for charge $$Q$$, right? Well, I don't know how.

 

[courteous]

I know that $$\overline{P}_{dissipated}\times t=\overline{P}_{supplied}\times t=3.60\text{ }mJ$$. How do I find $$V$$?

 

[Dmytry]

the energy stored in charged capacitor is v² * c / 2 .
You can find that formula if you note that V=Q/C and dE=V*dQ , and integrate charge from Q=0 to Q=Q₀ (you don't have to integrate by time, you can integrate by charge. Which makes more sense. Suppose I connected it to variable resistor that is being turned left and right by a monkey, the energy would of been the same, time doesn't matter there)

 

[courteous]

Thank you Dmytry!

 

[DeeNos]

Hello, based on the given information, we can use the formula for energy dissipated in a resistor as E = (1/2)CV^2. We also know that Q=CV, so we can rearrange the equation for energy to solve for V: V= sqrt(2E/C). Plugging in the given values, we get V= sqrt(2*3.60 mJ/2\mu F) = 0.12 V. Now, we can use this value of V in the equation Q=CV to find the charge Q: Q= (2\mu F)(0.12 V) = 120 \mu C. Therefore, the charge Q_0 is 120 \mu C.