Question

Find the charge on the capacitor after 1 sec of opening the switch at t = ∞?

• A. 20e⁻¹⁰ µC
• B. 25e⁻¹⁰ µC
• C. 30e⁻¹⁰ µC
• D. 35e⁻¹⁰ µC

Answer 1

Answer: (B)

25e⁻¹⁰ µC

Answer 2

To solve this problem, we need to follow two steps:

1. Calculate the initial charge on the capacitor (Q₀) at t = ∞ before opening the switch.
2. Calculate the charge on the capacitor after 1 second of opening the switch, considering the discharge process.

Step 1: Calculate the initial charge on the capacitor (Q₀) at t = ∞ (switch closed).

At steady state (t = ∞), a capacitor acts as an open circuit. This means no current flows through the branch containing the capacitor and the 5 kΩ resistor.

The circuit effectively becomes a series circuit with the 9V battery, the 12 kΩ resistor, and the 15 kΩ resistor.

The total resistance in this series circuit is:

$R_{total} = 12 \text{ k}\Omega + 15 \text{ k}\Omega = 27 \text{ k}\Omega$

The current flowing through this circuit is:

$I = \frac{V}{R_{total}} = \frac{9 \text{ V}}{27 \text{ k}\Omega} = \frac{9}{27 \times 10^3} \text{ A} = \frac{1}{3} \times 10^{-3} \text{ A} = \frac{1}{3} \text{ mA}$

The voltage across the 15 kΩ resistor is the voltage across the capacitor (since the capacitor branch is open, no voltage drop across 5 kΩ resistor).

$V_C(0) = I \times 15 \text{ k}\Omega = \frac{1}{3} \times 10^{-3} \text{ A} \times 15 \times 10^3 \Omega = 5 \text{ V}$

The initial charge on the capacitor is:

$Q_0 = C \times V_C(0) = 5 \mu\text{F} \times 5 \text{ V} = 25 \mu\text{C}$

Step 2: Calculate the charge on the capacitor after 1 second of opening the switch.

When the switch is opened, the 9V battery and the 12 kΩ resistor are disconnected from the circuit. The capacitor will now discharge through the remaining resistors.

Let's analyze the discharge circuit. The capacitor (C = 5 $\mu$F) is connected between two nodes. Let's call the node connected to the 15 kΩ resistor as Node A and the node connected to the 5 kΩ resistor as Node B. Both Node A and Node B are connected to ground through their respective resistors (15 kΩ and 5 kΩ).

To find the equivalent resistance ($R_{eq}$) for discharge, we need to determine the resistance seen by the capacitor across its terminals.

Let $V_C$ be the voltage across the capacitor, $V_A$ be the voltage at Node A, and $V_B$ be the voltage at Node B.

So, $V_C = V_A - V_B$.

Let $I_C$ be the current flowing out of the positive plate of the capacitor (Node A) and into the negative plate (Node B).

Current through 15 kΩ resistor ($R_1$) is $I_{R1} = V_A / R_1$.

Current through 5 kΩ resistor ($R_2$) is $I_{R2} = V_B / R_2$.

Applying KCL at Node A: $I_C + I_{R1} = 0 \implies I_C = -V_A/R_1$. (Assuming $I_C$ is current leaving A and entering C, so current leaving A into the circuit).

Applying KCL at Node B: $-I_C + I_{R2} = 0 \implies I_C = V_B/R_2$.

From these two equations, we have:

$-V_A/R_1 = V_B/R_2 \implies V_A R_2 = -V_B R_1 \implies V_A = -V_B \frac{R_1}{R_2}$

Substitute $V_A$ into $V_C = V_A - V_B$:

$V_C = -V_B \frac{R_1}{R_2} - V_B = -V_B \left(\frac{R_1}{R_2} + 1\right) = -V_B \frac{R_1 + R_2}{R_2}$

Now, solve for $V_B$:

$V_B = -V_C \frac{R_2}{R_1 + R_2}$

Substitute $V_B$ back into the expression for $I_C$:

$I_C = \frac{V_B}{R_2} = \frac{1}{R_2} \left(-V_C \frac{R_2}{R_1 + R_2}\right) = -\frac{V_C}{R_1 + R_2}$

For a discharging capacitor, the current is typically defined as $I_C = -\frac{dV_C}{dt}C$. The general form of discharge equation is $I = V/R_{eq}$.

Here, the current flowing out of the capacitor is $I_{discharge} = -I_C = \frac{V_C}{R_1 + R_2}$.

Thus, the equivalent resistance for discharge is $R_{eq} = R_1 + R_2$.

$R_{eq} = 15 \text{ k}\Omega + 5 \text{ k}\Omega = 20 \text{ k}\Omega$

The time constant ($\tau$) for the discharge is:

$\tau = R_{eq} \times C = 20 \times 10^3 \Omega \times 5 \times 10^{-6} \text{ F} = 100 \times 10^{-3} \text{ s} = 0.1 \text{ s}$

The charge on a discharging capacitor at time $t$ is given by:

$Q(t) = Q_0 e^{-t/\tau}$

We need to find the charge after 1 second ($t = 1 \text{ s}$):

$Q(1 \text{ s}) = 25 \mu\text{C} \times e^{-1 \text{ s} / 0.1 \text{ s}}$

$Q(1 \text{ s}) = 25 \mu\text{C} \times e^{-10}$

So, the charge on the capacitor after 1 second is $25e^{-10} \mu\text{C}$.