91Ó°ÊÓ

Figure 27-24

Use Kirchhoff’s loop law to find the differential equation and finding the solution of this differential condition and using the initial conditions, find the final equilibrium charge on the capacitor. For finding the time required, solve the equation for the charge of capacitor for 50% of charge.

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero.

Formulae are as follow:

V=IR

Where,

I is current, V is voltage, R is resistance.

The final charge on the capacitor for the given three sections:

For fig.1)

Apply Kirchhoff’s law to the first section to find the final charge on the capacitor C,

$\mathrm{Î\mu }-iR-\frac{q}{C}-iR=0$

$\mathrm{Î\mu }-2iR-\frac{q}{C}=0$

Substituting for current i=$\frac{dq}{dt}$

$\mathrm{Î\mu }=2R\frac{dq}{dt}+\frac{q}{C}…………………………………………………………………….1)$

The solution of this equation is given by,

$q=C\mathrm{Î\mu }(1-e^{-\frac{t}{2RC}})……………………………………………………………….2)$

Applying the initial conditions i.e. initially, the capacitor is uncharged.

q=0 When t=0

Note that, at t=0 the term$e^{-\frac{t}{2RC}}$is the unity.

So, the equation2) gives q=0

And as time goes to infinity, our capacitor charges completely.

That is, at$t=∞$, the second term of the equation 2) $e^{-\frac{t}{2RC}}$goes to zero and the equation 2) gives the final (equilibrium) charge on the capacitor.

$q=C\mathrm{Î\mu }……………………………………………………………………………..3)$

This is the final charge that appears on the capacitor after a long time.

For fig.2)

The resistors are connected in parallel. So, calculate the equivalent resistance.

$R_{eq}=\frac{R×R}{2R}$

$R_{eq}=\frac{R}{2}$

Now, replace these two resistors by the equivalent resistor.

This equivalent resistor and the capacitor will be in series and applying Kirchhoff law to this loop,

q=C$\mathrm{Î\mu }$

For fig.3)

Applying Kirchhoff’s loop law to the fig.3), after connecting it to the battery of emf, the differential equation is.

$R\frac{dq}{dt}=\mathrm{Î\mu }-\frac{q}{C}…………………………………………………………...7)$

And the solution of this equation is,

$q=C\mathrm{Î\mu }(1-e^{-\frac{t}{CR}})…………………………………………………….8)$

And we get the final equilibrium charge as,

q=C$\mathrm{Î\mu }$

Hence,the final charge on the capacitor for three sections are same.

The time required for the capacitor to reach 50% of its initial charge:

$q=q_0{e}^{-\frac{t}{RC}}$

Charge is equal to the 50% of its final charge. So ,

$0.5q_0=q_0{e}^{-\frac{t}{RC}}$

$\ln (0.5)=-t/RC$

t=0.7RC

For the first case, total resistance would be,

$R_{eq1}=R_1+R_2=2R$

For second case, it would be equal to,

$\frac{1}{R_{eq2}}=\frac{1}{R_1}+\frac{1}{R_2}$

$R_{eq}=\frac{R}{2}$

For third case, we would have

$R_{eq3}=R$

From the above equations, it can be seen that the time required for the capacitor to reach 50% of its final charge would be greatest for 1^st then 3^rd and the smallest for 2^nd.

Hence, the time required for the capacitor to reach 50% of its final charge is 1$>3>2$.

Therefore, find the total charge on the capacitor and the time required for 50% charging of the capacitor using the equation of Kirchhoff’s loop rule and solving the differential equation.