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- Note
- Things to remember

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 ‘V_{C}’ and ‘V_{R}’ 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 Q_{o} 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.

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 'Q_{o}' 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 ‘V_{C}’ and ‘V_{R}’ 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 log_{e} 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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## John

How t=CR?

Mar 17, 2017

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## John

What is time constant

Mar 17, 2017

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