Example 15.
1 kΩ
1 kΩ
t = 1 ms
b
a
6 kΩ
t=0
1 mA
1 kΩ 0.5 µF
12 kΩ
30 V
The switch is in position a for a long time before it is flipped to position b at t=0., where
it remains 1ms delay, the switch is moved back to position a.
a) Derive the expression for the voltage across the capacitor as t>0.
b) Plot the capacitor voltage versus time.
At the instant time when the switch is moved to position b, the capacitor voltage is 1V.
vc (0  )  vc (0  )  1V V0  1V
If the switch was to remain in position b, the capacitor would eventually charge to 20V.
V f  20V
The time constant of the circuit when the circuit is in position b is
  RC  4000  0.5  10 6  0.002 s
Therefore the capacitor voltage 0  t  1ms
vc (t )  V f  (V0  V f )e

t

 20  (1  20 )e 500t  20  19 e 500t V,
Note that the switch remains in position b for only 1 ms, this expression is valid only for
the time interval from 0 to 1 ms.
After the switch has been in position b for 1 ms, the voltage on the capacitor will be
vc (1ms)  20  19e 5000.001  8.48V t  1ms
There fore, when the switch is moved to position a again, the initial voltage on the
capacitor is vc (1ms )  8.48V . With the switch in position a, the final value of the
voltage should be 1V.
The time constant is   RC  2000  0.5  10 6  0.001 s
So the voltage is vc (t )  V f  (V0  V f )e
This expression is valid for t  1ms

t

 1  (8.48  1)e 1000 (t 0.001) t  1ms
v c (t(V)
)
v c (t )  20  19e 500t ,0  t  1ms
v c (t )  1  7.48e 1000 (t 0.001) , t  1ms
(ms
t
)