Question

The switch in the circuit in the figure is in position P for a long time and then moved to position Q at time \( t = 0 \). The value of \( \frac{d v(t)}{dt} \) at \( t = 0^+ \) is 

• A. 0 V/s
• B. 3 V/s
• C. -3 V/s
• D. -5 V/s

Hint:

For circuits involving capacitors, the rate of change of the voltage is related to the current through the capacitor. Use KVL and the capacitor voltage-current relationship to solve for \( \frac{d v(t)}{dt} \).

Answer 1

The Correct Option is C

We are given an RC circuit where the switch has been in position P for a long time, and at \( t = 0 \), it is moved to position Q. We are tasked with finding the value of \( \frac{d v(t)}{dt} \) at \( t = 0^+ \). Step 1: Analyze the behavior of the circuit.
In the long term (when the switch has been in position P for a long time), the capacitor in the circuit will be fully charged. At \( t = 0^+ \), when the switch is moved to position Q, the voltage across the capacitor will attempt to change. The rate of change of the voltage across the capacitor is related to the current through the capacitor by the following equation: \[ \frac{d v(t)}{dt} = \frac{I(t)}{C} \] where \( I(t) \) is the current through the capacitor and \( C \) is the capacitance. Step 2: Use Kirchhoff’s Voltage Law (KVL).
By applying KVL to the circuit and considering the initial conditions, we can find the rate of change of the voltage across the capacitor at \( t = 0^+ \). Solving the equations gives us the value \( \frac{d v(t)}{dt} = -3 \, \text{V/s} \).
Thus, the correct answer is option (C). Final Answer: -3 V/s