Note on Charging and Discharging Capacitor

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Charging a Capacitor through Resistance

(a) Discharging of a capacitor. (b) Graph of discharging nature of a capacitor.
(a) Discharging of a capacitor. (b) Graph of discharging nature of a capacitor.

A capacitor is always charged through a resistance in order to control the flow of electric charge.

Suppose a capacitor having capacitance ‘C’ is being charged through resistance ‘R’ by using a cell of terminal potential difference ‘V’. Consider at any instant of time ‘t’. The charge present in capacitor ‘Q’ and potential difference across capacitor and resistance be ‘VC’ and ‘VR’ respectively.

Then,

$$V = V_R + V_C \dots (i)$$

Also \(Q =C.V_C \)

$$\therefore V_C = \frac QC \dots (ii)$$

And from Ohm’s law;

$$V_R = IR$$

$$V_R = \frac {dQ}{dt} R \dots (iii)$$

Where, \(I = \frac {dQ}{dt}\) is instantaneous current. If Qo be the maximum charge hat can be stored in capacitor is fully charged

$$\therefore Q_o = C.V$$

$$\text {or,} V = \frac {Q_o}{C} \dots (iv) $$

Using equation (ii), (iii) and (iv), we get in equation (i)

$$\text {or,} \frac {Q_o}{C} = \frac {Q}{C} + \frac {dQ}{dt} R$$

$$\text {or,} \frac {1}{C} {(Q_o -Q)} = \frac {dQ}{dt} R $$

$$\text {or,} \frac {1}{CR} dt = \frac {dQ}{Q_o - Q} $$

At time t = 0, charge present in capacitor is 0 and at time 't', charge present is 'Q'. So, integrating above expression without suitable limits. We get

$$\text {or,} \int _0^t \frac{1}{CR} dt = \int_0^Q \frac {dQ}{Q_o - Q}$$

$$\text {or,}\frac {1} {CR}\int _0^t dt =\int_0^Q \frac {dQ}{Q_o - Q}$$

$$\text {or,}\frac {1} {CR} [t]_0^t = \left [ -\log_e (Q_o - Q) \right ]_0^Q $$

$$\text {or,}\frac {1} {CR} [t - 0] =-\log_e (Q_o - Q) + \log_e (Q_o - 0) $$

$$\text {or,}-\frac {t} {CR} = -\log_e\left (\frac { Q_o - Q} {Q_o}\right )$$

$$\text {or,} \log_e\left (\frac { Q_o - Q} {Q_o}\right )= - \frac {t}{CR} $$

$$\text {or,}\frac { Q_o - Q} {Q_o} = e^{- \frac {t}{CR} }$$

$$\text {or,}Q_o - Q =Q_o e^{- \frac {t}{CR} }$$

$$\text {or,}Q_o - Q_o e^{- \frac {t}{CR}} = Q$$

$$\therefore Q =Q_o (1 - e^{- \frac {t}{CR}}) $$

This is the expression for charge stored in a capacitor at any instant 't'

Time constant

For t = CR

$$Q =Q_o (1- e^{- 1}) $$

$$=Q_o (1- \frac 1e) $$

$$= Q_o (1 - 0.368)$$

$$= Q_o \times 0.632$$

$$\boxed {\therefore Q = 63.2 \%of Q_o}$$

Hence, the time constant or relaxation time for charging a capacitor is given by the product of capacitance and resistance connected in series is defined as:

The time required to charge the capacitor up to 63.2% of the total charge that it can store.

Discharging a Capacitor through Resistance

(a) Charging a Capacitor (b) Graph of charging nature of capacitor
(a) Charging a Capacitor (b) Graph of charging nature of capacitor

During the discharge process of a charged capacitor, in order to control the flow of charge a resistance is used.

Consider a capacitor (C), fully charged with charge 'Qo' is being discharged through resistance 'R' and at any instant of time 't' the charge stored in a capacitor 'Q' and the potential difference across capacitor and resistance are ‘VC’ and ‘VR’ respectively. Then,

$$V_C = V_R \dots (i)$$

But, \(Q = C. V_C\)

$$\therefore V_C = \frac QC \dots (ii)$$

and \(V_R = IR\)

$$\therefore V_R = -\frac {dQ}{dt}R \dots (iii)$$

where \(I = -\frac {dQ}{dt}\) is the value of current at that instant and -ve sign indicates xharge decreases with increase in time.

Using equation (ii) and (iii) in equation (i), we get

$$\text {or,} \frac QC= -\frac {dQ}{dt} R$$

$$\text {or,} \frac {1}{CR}= -\frac {dQ}{dt} \times \frac {1}{Q}$$

$$\text {or,} -\frac {1}{CR} dt = \frac {dQ}{Q} $$

Integrating between suitable limits. We get,

$$\text {or,}- \frac {1}{CR} \int _o^t dt = \int _{Q_o}^Q\frac {dQ}{Q} $$

$$\text {or,}- \frac {1}{CR} [t]_0^t = [\log_e Q]_{Q_o}^Q $$

$$\text {or,}- \frac {1}{CR} [t - 0] = \log_e Q - \log_e Q_o $$

$$\text {or,}- \frac {t}{CR} = \log_e\left (\frac {Q}{Q_o} \right) $$

multiplying by loge on both sides

$$\text {or,}e^{- \frac {t}{CR}} =\frac {Q}{Q_o} $$

$$ \therefore Q = Q_oe^{- \frac {t}{CR}} $$

This is the expression for charge present in a capacitor during the process of discharge at any time 't'. It indicates the charge stored decreases exponentially.

Time constant

For t = CR

We have,

$$Q = Q_o e^{-1}$$

$$= Q_o \left (\frac 1e \right)$$

$$= Q_o (0.368) $$

$$\boxed {\therefore Q = 16.8 \% of Q_o}$$

Hence, the time for the discharge circuit is defined as the time constant (t = CR) discharges quickly and a circuit having large time constant requires more time to charge.

 A capacitor is always charged through a resistance in order to control the flow of electric charge.

The time required to charge the capacitor up to 63.2% of the total charge that it can store.

The time for the discharge circuit is defined as the time constant (t = CR) discharges quickly and a circuit having large time constant requires more time to charge.

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Very Short Questions

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John

How t=CR?


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John

What is time constant


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