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## Q. 5.4

Capacitor Voltage

The switch in the circuit shown in Figure 5.16 has stayed in position 1 for a long time period. At time t = 0 the switch moves to position 2.

a. Find $v_1 (0^-)$, that is, $v_1 (t)$ prior to the displacement of the switch.

b. Find $v_2 (0^+)$, that is, $v_2 (t)$ right after the displacement of the switch.

c. Find an expression for $v_C (t)$ for t > 0.

d. Find the value of t at which the capacitor voltage reduces to 6 V.

## Verified Solution

a. After a long time period of the switch being in position 1, the capacitor is fully charged and can be replaced by an open circuit. Thus, using voltage divider rule, the voltage across the 16-kΩ resistor is:

$v_1\left(0^{-}\right)=v_{C}\left(0^{-}\right)=15 \frac{16}{16+8}=10 V$

In other words, the fully charged capacitor has been charged to 10 V.

b. After the movement of the switch to position 2, $v_{C}\left(0^{+}\right)=v_{C}\left(0^{-}\right)=10 V$. Thus,

$v_2\left(0^{+}\right)=10 V$

c. At t = 0, when the switch is changed to position 2, the capacitor starts discharging through the 2-kΩ resistor. Using the same procedure as in the two previous examples, the voltage across the capacitor at time t = 0 corresponds to:

$v_{i}=v_{C}\left(0^{-}\right)=v_{C}\left(0^{+}\right)=10 V$

For t > 0, the time constant is:

$\tau=R C=2 \times 10^3 \times 1.5 \times 10^{-6}=3 \times 10^{-3} s$

Substituting $v_{i}$ and τ in Equation (5.23):

$v_{C}(t)=v_{i} e^{-\frac{t}{\tau}}$     (5.23)

$v_{C}(t)=10 e^{-\frac{t}{3 \times 10^{-3}}}$

d. Substituting $v_{C}(t)=6 V$:

$6=10 e^{-\frac{t}{3 \times 10^{-3}}} \rightarrow e^{-\frac{t}{3 \times 10^{-3}}}=0.6 \rightarrow-\frac{t}{3 \times 10^{-3}}=\ln 0.6$

The value of t at which the capacitor voltage drops to 6 V is t = 1.5325 ms.