Question

A mega ohm resistor and an uncharged $1\mu F$ capacitor are connected in a single loop circuit with a constant sources of $4$ volt.At one second after the connection is made what are the rates at which energy is being stored in the capacitor:

• A. $\frac{16}{3}(1-e^{-1/3})e^{-1/3}\mu J/s$
• B. $\frac{16}{3}(1-e^{-2/3})\mu J/s$
• C. $\frac{16}{3}{e}^{-2/3}\mu J/s$
• D. None of these

Answer 1

The correct option is A $\frac{16}{3}(1-e^{-1/3})e^{-1/3}\mu J/s$

Energy is being stored in the capacitor is $\frac{1}{2}\frac{q_0^2}{c}{(1-e^{-\frac{t}{Rc}})}^2$

$E=\frac{1}{2}\left(\frac{16}{c}\right){(1-e^{-\frac{t}{Rc}})}^2$

So, $Rate=\frac{dE}{dt}=\frac{16}{2c}\times 2(1-e^{-\frac{t}{Rc}})\left(\frac{1}{Rc}\right)e^{-\frac{t}{Rc}}$

$=\frac{16}{3}(1-e^{-\frac{t}{Rc}})e^{-\frac{t}{Rc}}$

If $t=1s$ then,

$\frac{dE}{dt}=\frac{16}{3}(1-e^{-\frac{1}{3}})e^{-\frac{1}{3}}$