D.C. Transients in RC Circuit

WORKED EXAMPLES

2. Case 2: (a) R-C Transients (Charging)

Consider the R-C series combination. A DC voltage E is applied to this combination through the switch as shown in fig. 3.13.

Let Q₀ = initial charge on the capacitor. So, the initial voltage on the capacitor = V₀ =  Q₀ / C.Let the polarities of Vo be as shown:

After 1 seconds of closing the switch, applying Kirchoff's voltage law to the circuit we get,

 [Note: If the polarities of Vo are opposite to that shown in the figure, then replace Vo in the above equation by -V_o]

In the present case let us assume that the initial charge on the capacitor is 0. Hence V1 = 0. Substituting V_o = 0 in equation (iii) we get, the final equation as

Taking Laplace transform on both sides, we get

E / R is the initial value of the current denoted by I, RC is called time constant denoted by λ.

Therefore, we can express the charging current as,

i (t) = I e ^–(t/λ) ... (vi)

This charging current seems to be exponentially decreasing as is observed from the equation (vi).

The initial current = the value of i when (t = 0) = I

The steady current = the value of i when (t = ∞) = 0

The variation of the current with time is graphically shown in the fig. 3.14.

The transient voltages e_R and e_C are obtained as below:

e_R exponentially decreases from E to 0 as t increases from 0 to ∞.

e_C exponentially increases from 0 to E as t increases from 0 to ∞.

These observations are made from equations (vii) and (viii).

Graphically, the variations of e, and ec can be plotted as shown in the fig. 3.15.

Definition of Time - Constant (λ) for R-C circuit

In the equation (viii), substitute that λ = t.

Then i = I × e^-1 = 0.368 × I

= 36.8% of the initial current

Thus, the time constant for the RC circuit is the time during which the current falls to 36.8% of initial current.

[Note: We can also get e by the formula 

3. Case (b) R - C Decaying Transient (Discharging)

Consider the circuit of the fig. 3.16. The switch has been in position 1 for sufficient time to establish steady state condition. Let this switch be moved to position 2 at the instant t = 0.

Under steady state condition, the capacitor is fully charged. The voltage across C is E with the polarities as shown in the fig.3.16, The differential equation of the circuit is,

Taking Laplace transform and rearranging the terms, we get

This current is called discharging current.

The current curve is as shown in the fig. 3.16.

WORKED EXAMPLES

Example 7 A resistance R and a 2uF capacitor are connected in series across a 200V direct supply. Across the capacitor is a neon lamp that strikes at 120V. Calculate R to make the lamp strike 5 sec after the switch has been closed. If R = 5M 2, how long will it take the lamp to strike?

Solution: Case (i):

Example 8 A resistor is connected across the terminals of a 20 uF capacitor which has been previously charged to a.p.d. of 500V. If the p.d. falls to 300V in 0.5 min, calculate the resistance in megaohms.

Solution :

This is case of discharging of a capacitor.

The p.d. across C at any time t after the switch is closed is,

We can proceed to solve this equation by Laplace transform method. It yields the final solution which we already obtained.]

Example 9 Find how long it takes after the key is closed before the total current from the supply reaches 25 mA, when V = 10V, R1 = 500 2, R2 = 700 2 and C = 100 μF.

Solution:

Let i1 = the current through R1

i2 = current (charging current)

through the capacitor after t seconds of closing the switch.

I1 = V / R1 = 10 / 500 A = 20 mA

It is constant.

Given that Total current = i = 25 mA

By applying KCL,

Substituting the known values in equation (i), we get