# Chapter 8: D.C. Transients¶

## Example 8.1, 253¶

In [1]:
#Variable declaration
C = 8e-06;    # Value of capacitance of capacitor, farad
R = 0.5e+06;    # Value of series resistor, ohm
E = 200;    # Value of d.c. voltage supply, volt

#Calculations&Results
# Part (a)
tau = C*R;    # Time constant of the R-C circuit while charging, s
print "The circuit time constant while charging = %1d s"%tau

# Part (b)
I_0 = E/R;    # Initial charging current through capacitor, A
print "The initial charging current through capacitor = %3d micro-ampere"%(I_0/1e-06);

# Part (c)
t = 4;  # Time after the supply is connected, s
v_C = 0.632*E;  # p.d. across the capacitor 4s after the supply is connected, V
v_R = E - v_C;  # p.d. across the resistor 4s after the supply is connected, V
print "The p.d. across resistor and capacitor %d s after the supply is connected = %5.1f V and %4.1f V respectively"%(t, v_C, v_R);

The circuit time constant while charging = 4 s
The initial charging current through capacitor = 400 micro-ampere
The p.d. across resistor and capacitor 4 s after the supply is connected = 126.4 V and 73.6 V respectively


## Example 8.2, Page 255¶

In [2]:
#Variable declaration
C = 0.5e-06;    # Value of capacitance of capacitor, farad
R1 = 220e+03;    # Value of series resistor, ohm
R2 = 110e+03;    # Value of parallel resistor, ohm
E = 150;    # Value of d.c. voltage supply, volt

#Calculations&Results
# Part (a)
tau = C*R1;    # Time constant of the R1-C circuit while charging, s
print "The circuit time constant while charging = %4.2f s"%tau
I_0 = E/R1;    # Initial charging current through capacitor, A
print "The initial charging current through capacitor = %3d micro-ampere"%(I_0/1e-06)

# Part (b)
tau = C*(R1+R2);    # Time constant of the R1-C-R2 circuit while discharging, s
print "The circuit time constant while discharging = %4.2f s"%tau
I_0 = E/(R1 + R2);    # Initial discharging current through capacitor, ampere
i = 0.368*I_0;    # Discharge current after one time constant, ampere
V_R2 = i*R2;    # Potential difference across R2 after one time constant, volt
print "The p.d. across R2 after one time constant while discharging = %4.1f volt"%V_R2

The circuit time constant while charging = 0.11 s
The initial charging current through capacitor = 681 micro-ampere
The circuit time constant while discharging = 0.16 s
The p.d. across R2 after one time constant while discharging = 18.4 volt


## Example 8.3, Page 258¶

In [3]:
#Variable declaration
E = 110.;    # Value of d.c. voltage supply, volt
L = 1.5;    # Inductor value, henry
R = 220;    # Value of series resistor, ohm

#Calculations&Results
# Part (a)
di_dt = E/L;    # The initial rate of change of current through inductor, H
print "The initial rate of change of current through inductor = %5.2f A/s"%di_dt

# Part (b)
I = E/R;    # The final steady current, A
print "The final steady current through inductor = %3.1f A"%I

# Part (c)
tau = L/R;    # The time taken for the current to reach its fi nal steady value, s
print "The time taken for the current to reach its final steady value = %4.1f ms"%(5*tau/1e-03);

The initial rate of change of current through inductor = 73.33 A/s
The final steady current through inductor = 0.5 A
The time taken for the current to reach its final steady value = 34.1 ms