Question

The key \( S_1 \) is closed and \( S_2 \) is open. The value of current in the resistor after 5 seconds is:

• A. \( \frac{1}{2} \, \text{mA} \)
• B. \( \sqrt{2} \, \text{mA} \)
• C. \( \frac{1}{\sqrt{e}} \, \text{mA} \)
• D. \( \frac{1}{2} \, \text{e mA} \)

Hint:

The current in an RC circuit decreases exponentially as the capacitor charges, and this can be described by \( I(t) = \frac{V}{R} e^{-\frac{t}{RC}} \).

Answer 1

The Correct Option is C

To determine the current in the resistor 5 seconds after the switch configuration, we consider that initially, the key \( S_1 \) is closed and \( S_2 \) is open. The circuit is likely an RC (resistor-capacitor) circuit due to the context provided.

In an RC circuit, the current \( I(t) \) flowing through the circuit when charging or discharging a capacitor is described by the following equation:

\[ I(t) = I_0 \cdot e^{-\frac{t}{RC}} \]

where \( I_0 \) is the initial current, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is the time.

The problem implies we need the current after a certain time has passed, specifically at \( t = 5 \) seconds.

Given that the answer is in terms of \( \frac{1}{\sqrt{e}} \), we evaluate the expression at \( t = 5 \) assuming the effective time constant \( \tau = RC \) simplifies this way:

\[ I(5) = I_0 \cdot e^{-\frac{5}{\tau}} \]

For simplicity and from the context of the correct answer being \( \frac{1}{\sqrt{e}} \, \text{mA} \), this suggests that:

\[ e^{-\frac{5}{\tau}} = \frac{1}{\sqrt{e}} \]

Solving for \( \tau \), this indicates:

\[ -\frac{5}{\tau} = -\frac{1}{2} \Rightarrow \tau = 10 \]

Therefore, the time constant \( \tau = RC = 10 \) seconds fits the scenario presented.

Thus, the current in the resistor after 5 seconds is indeed:

\[ I = \frac{I_0}{\sqrt{e}} \, \text{mA} \]

This matches the correct answer, \( \frac{1}{\sqrt{e}} \, \text{mA} \).