Diff. Eq. for charging of Capacitor!

Electricity and Magnetism Level 3

The differential equation of charging of a capacitor is as given below:

$\large \frac{1}{K_1} \frac{dq}{dt} + K_2q = K_3$

The time constant $\large \tau$ and steady state charge $\large q_0$ are respectively:

None of These

\frac{1}{K_1}; K_3

\frac{1}{K_1K_2}; \frac{K_3}{K_1}

\frac{K_2}{K_1}; K_2K_3

\frac{1}{K_1K_2}; \frac{K_3}{K_2}

1 solution

Tanishq Varshney
Jun 13, 2015

basic equation for charging of capacitor in a $RC$ circuit is

$\large{\frac{q}{C}+IR=E}$

where, $E$ is emf of cell/ battery , $C$ is capacitance and $R$ is resistance.

$\large{\frac{q}{C}+\frac{dq}{dt}R=E}$

comparing with this equation we get

$C=\frac{1}{K_{2}}$ and $R=\frac{1}{K_{1}}$

$\large{\tau=RC=\frac{1}{K_{1}K_{2}}}$

For steady state $q=q_{o}$ and $\frac{dq}{dt}=0$

$\large{q_{0}=\frac{K_{3}}{K_{2}}}$

@Nishant Rai can u plz reply to my solution of youngs modulus problem

Tanishq Varshney - 6 years ago

Was it excellent or there was any problem can you provide a link.

Satvik Choudhary - 6 years ago

@satvik choudhary what are you trying to say?

Here is the link to the problem Tanishq's asking for - Calculate Young's Modulus!

Nishant Rai - 6 years ago