+4 a Ams ltt von
QS. The circuit below contains a switch that has been closed for a long time. [20 pts]
Att =0, the switch was opened. The initial and final current/voltage
values were measured b

Question

+4 > a Ams ltt von
QS. The circuit below contains a switch that has been closed for a long time. [20 pts]
Att =0, the switch was opened. The initial and final current/voltage
values were measured by another EGRE 206 student (shown in the table below).
2H
— fi
i +
s0 Ve =325mF [3 1=0 [v(0) [ 00) [ 140) [ vetoo) [fem) |
il 15v [2a [2a [ sv | oa |
100
ie 150
i" wm 3Y
il WR) AREER | |
LL a) Indicate whether this circuit is overdamped or underdamped. Explain why, |
2 b) Determine the expression for ve(t)for t > 0, |
Find the capacitor value needed to make the circuit critically damped, leaving
all of the other components unchanged.
I&

VIPSolutions · Accepted Answer

## Solution

### a) Damping Analysis

To determine whether the circuit is overdamped or underdamped, we need to evaluate the damping ratio ($\zeta$) of theLC circuit formed by the inductor2H), the resistor (15), and the capacitor (31.mF).

The formula for the ratio is:

$$
\zeta =frac{R}{2} \{\frac{C}{L}}
$$

Where:
- $R = 15 \, \Omega$
- $C = 31.25 \, \text{mF} = 31.25 \times 10^{-3} \, \text{F}$
- $L = 2 \, \text{H}$

Substitute the values:

$$
\zeta = \frac{15}{2} \sqrt{\frac{31.25 \times 10^{-3}}{2}}
$$

Calculate $\zeta$. If $\zeta > 1$, the circuit is overdamped. If $\zeta < 1$, it is underdamped.

### b) Expression for $v_c(t)$ for $t > $

The expression for $v_c(t)$ in an RLC circuit can be found using the standard form of the second-order differential equation solution. For an underdamped system:

$$
v_c(t) = V_f + (V_ - V_f) e^{-\alpha t} \cos(\omega_d t + \phi)
$$

Where:
- $V_f = 5 \, \text{V} $ (final voltage)
- $V_ = 15 \, \text{V} $ (initial voltage)
- $\alpha = \frac{R}{2L}$
- $\omega_d = \omega_n \sqrt{1 - \zeta^2}$
- $\omega_n = \frac{1}{\sqrt{LC}}$

Calculate $\alpha$, $\omega_n$, and $\omega_d$ using the given values.

### c) Capacitor Value for Critical Damping

For critical damping, the damping ratio $\zeta = 1$. The condition for critical damping is:

$$
R = 2 \sqrt{\frac{L}{C'}}
$$

Solve for $C'$:

$$
C' = \frac{L}{\left(\frac{R}{2}\right)^2}
$$

Substitute the given values:

$$
C' = \frac{2}{\left(\frac{15}{2}\right)^2}
$$

Calculate $C'$ to find the capacitor value needed for critical damping.

VIPSolutions · Answer

# Circuit Analysis

## a) Damping Analysis

To determine whether the circuit is overdamped or underdamped, we evaluate the damping ratio ($\zeta$) of the RLC circuit formed by the following components:

- Inductor ($L = 2 \, \text{H}$)
- Resistor ($R = 15 \, \Omega$)
- Capacitor ($C = 31.25 \, \text{mF} = 31.25 \times 10^{-3} \, \text{F}$)

The formula for the damping ratio is:

$$
\zeta = \frac{R}{2} \sqrt{\frac{C}{L}}
$$

**Substituting the values:**

$$
\zeta = \frac{15}{2} \sqrt{\frac{31.25 \times 10^{-3}}{2}}
$$

**Calculation Steps:**
1. Calculate $\frac{C}{L} = \frac{31.25 \times 10^{-3}}{2}$.
2. Take the square root and multiply by $\frac{R}{2}$.

- If $\zeta > 1$, the circuit is overdamped.
- If $\zeta < 1$, it is underdamped.

## b) Expression for $v_c(t)$ for $t > 0$

The expression for the capacitor voltage $v_c(t)$ in an RLC circuit can be described by the equation:

$$
v_c(t) = V_f + (V_0 - V_f) e^{-\alpha t} \cos(\omega_d t + \phi)
$$

Where:
- $V_f = 5 \, \text{V}$ (final voltage)
- $V_0 = 15 \, \text{V}$ (initial voltage)
- $\alpha = \frac{R}{2L}$
- $\omega_d = \omega_n \sqrt{1 - \zeta^2}$
- $\omega_n = \frac{1}{\sqrt{LC}}$

**Calculating Parameters:**

1. **$\alpha$**:
   $$
   \alpha = \frac{15}{2 \times 2} = \frac{15}{4} = 3.75
   $$

2. **$\omega_n$**:
   $$
   \omega_n = \frac{1}{\sqrt{2 \times 31.25 \times 10^{-3}}} 
   $$

3. **$\omega_d$**:
   Substitute $\zeta$ to find $\omega_d$.

## c) Capacitor Value for Critical Damping

For critical damping ($\zeta = 1$), the condition is:

$$
R = 2 \sqrt{\frac{L}{C'}}
$$

**Solving for $C'$**:

$$
C' = \frac{L}{\left(\frac{R}{2}\right)^2}
$$

**Substituting the values**:

$$
C' = \frac{2}{\left(\frac{15}{2}\right)^2}
$$

**Calculating $C'$**:

1. Find $\frac{15}{2} = 7.5$.
2. Square it: $7.5^2 = 56.25$.
3. Finally, compute $C' = \frac{2}{56.25}$.

This value gives the required capacitor for critical damping.

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