The maximum capacitance of a variable capacitor is $33\; pF$. What should be the self-inductance to be connected to this capacitor for the natural frequency of the LC circuit to be $810 \; KHz$. Corresponding to A.M. broadcast band of Radio Nepal?

Given, Capacitance $(C) = 33\; pF = 33 * 10^{-12}\;F$

Let, $L$ be the self-inductance connected with the capacitance, so that the natural frequency of this LC circuit is $810\; KHz$,

$i.e\; \; \; \;  f = 810 \; * 10^3 \; Hz $

Now, we have 

$f = \frac{1}{2 \pi \sqrt{LC}}$

$L =$ $ \frac{1}{4 \; \pi^2 \; (810 \; * \;10^3)^2 \; * \; 33 \; * \; 10^{-12}}$  $= 1.17 * 10^{-3}\; H$

Thus, Inductance connected $(L) = 1.17 * 10^{-3}\;H$

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Capacitor's and Dielectrics | PPT Slide

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4.4 Discharging of a Capacitor through Resistor:

To discharge a capacitor through a resistor, let us consider a capacitor of capacitance $(C)$ is initially charged to a potential difference $V_{0}$ and charge $q_{0}$, then

$q_{0} = C V_{0}$ .......... (i)

Now, the charged capacitor is joined to a resistor of resistance $(R)$ in series as shown in figure.

Figure 1: Circuit diagram for discharging Capacitor

At a time $(t)$ after the discharging through $(C)$ has begun, let the $V_{C}$ is the potential across the capacitor and $q$ be the charge. If current $(I)$ flows through $R$, then we have

$V_{C} = V_{R}$ .......... (ii)

or, $\frac{q}{c}$ = $IR$

or, $\frac{q}{c}$ = $-\frac{dq}{dt}$$R$    [∵ $I$ = $-\frac{dq}{dt}$;  '-$ve$' sign shows that $q$ decreases with increasing $t$]

or, $\frac{dq}{q}$ = $-\frac{1}{RC}$

Integrating both sides, we get

or, $\int_{q}^{q_{0}} \frac{dq}{q} = \int_{0}^{t} - \frac{1}{RC}$

or, $[\ln q]$$_{q_{0}}^{q}$ = $- \frac{1}{RC}$$[t]$$_{0}^{t}$

or, $\ln q - \ln q_{0}$ = $-\frac{t}{RC}$

or, $\ln$ ($\frac{q}{q_{0}}$) = $- \frac{t}{RC}$

or, $\frac{q}{q_{0}}$ = $e^{-\frac{t}{RC}}$

∴ $q = q_{0}$$ e^{-\frac{t}{RC}}$ .......... (iii)

This is called the decay of charge equation. Clearly, $q$ decreases exponentially with time (t) as shown in figure.

Figure 2: Variation of charge with time in a discharging Capacitor

If t = RC in above equation, then from the equation (iii), we get

or, $q = q_{0}$$ e^{-\frac{RC}{RC}}$ = $q_{0}\;e^{-1}$ =  $\frac{q_{0}}{e}$ $= 0.37\; q_{0} = 37 \%$ of $ q_{0}$

Thus the discharging time constant may be defined as the time at which the charge on the capacitor during discharging becomes about 37 % of the initial charge.

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4.4 Charging of a Capacitor through a resistor:

Let us consider a capacitor of capacitance $(C)$ and a resistor of resistance $(R)$ are connected in a series with a potential $(V)$, as shown in the figure.

Figure 1: Circuit diagram for a charging Capacitor

Initially, there is no charge in the capacitor. i.e. at time $t = 0$, the charge in the capacitor $C$ is also zero.

After a time $(t)$ sec, the charge on a capacitor is $q$. Let the current in the circuit is $I$, with a potential difference across the capacitor is $V_{C}$, and the potential difference across the resistor is $V_{R}$. Then,

$V{_C}$ = $\frac{q}{c}$ $\;\;\;$ and $\;\;\;$ $V_{R}$ = $IR$

If $q_{0}$ is the maximum charge stored in the capacitor, then

$q_{0}$ = $CV$

From Figure, $V$ = $V_{C}$ + $V_{R}$ .......... (i)

or, $\frac{q_{0}}{C}$ = $\frac{q}{C}$ + $IR$

or, $\frac{q_{0}- q}{C}$ = $R$ $\frac{dq}{dt}$

or, $\frac{dq}{q_{0}- q}$ = $\frac{1}{RC}$$dt$ .......... (ii)

Integrating equation (ii), we get

or, $\int_{0}^{q}\frac{dq}{q_{0}- q}$ = $\int_{0}^{t}\frac{1}{RC}$$dt$

or, $[$-$\;\ln(q_{0}-q)$]$_{0}^{q}$ = $\frac{1}{RC}[t]_{0}^{t}$

or, - $\ln(q_{0} - q)$ + $\ln q_{0}$ = $\frac{t}{RC}$

or, $\ln(q_{0} - q) - \ln q_{0}$ = $- \; \frac{t}{RC}$

or, $\ln$$\frac{q_{0} - q}{q_{0}}$ = $- \frac{t}{RC}$

or, $\frac{q_{0} - q}{q_{0}}$ = $e^{-\frac{t}{RC}}$

or, $q_{0} - q$ = $q_{0}\;$$e^{-\frac{t}{RC}}$

or, $q$ = $q_{0}( 1$ - $e^{-\frac{t}{RC}})$ .......... (iii)

A graph between charge $(q)$ and time $(t)$ during charging of a capacitor as shown in figure.

Figure 2: Variation of charge with time in a charging capacitor 

If $t = RC$, then from equation (iii) we get,

or, $q$  =  $q_{0}( 1 -$ $e^{-\frac{RC}{RC}}$$)$  =  $q_{0}(1 -$ $e^{-1}$$)$

or, $q$ = $q_{0}(1 - 0.37)$

or, $q$ = $q_{0} * 0.63$

or, $q$ = $63 \; \%$ of $q_{0}$ .......... (iv)

This is the equation for growth of charge.

∴ The charging time constant (or $R-C$ time constant) of a capacitor is defined as the time interval in which the capacitor charges by about $63 \%$ of its maximum charge.

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