Question

A signal $V_M = 5\sin(\pi t/3) V$ is applied to the circuit consisting of a switch S and capacitor $C = 0.1 \mu F$, as shown in the figure. The output $V_x$ of the circuit is fed to an ADC having an input impedance consisting of a $10 M\Omega$ resistance in parallel with a $0.1 \mu F$ capacitor. If S is opened at $t = 0.5 s$, the value of $V_x$ at $t = 1.5 s$ will be ________ V (rounded off to two decimal places).
Note: Assume all components are ideal.

 

Hint:

The voltage across a capacitor cannot change instantaneously. When a switch opens or closes, the initial voltage across the capacitor is maintained. The subsequent behavior is governed by the time constant of the RC circuit.

Answer 1

Correct Answer: 1.5

Step 1: Determine the voltage across the capacitor $C$ at $t = 0.5 s$.
$V_M(0.5) = 5\sin(\pi \times 0.5 / 3) = 2.5 V$. $V_c(0.5^-) = 2.5 V$, and due to continuity, $V_c(0.5^+) = 2.5 V$. 
Step 2: Determine the equivalent capacitance after the switch opens.
$C_{eq} = C + C_{ADC} = 0.1 \mu F + 0.1 \mu F = 0.2 \mu F$. 
Step 3: Determine the discharge time constant.
$\tau = R_{ADC} C_{eq} = (10 \times 10^6) \times (0.2 \times 10^{-6}) = 2 s$. 
Step 4: Write the voltage across the equivalent capacitance as a function of time for $t \ge 0.5 s$.
$V_x(t) = V_x(0.5) e^{-(t - 0.5) / \tau} = 2.5 e^{-(t - 0.5) / 2}$. 
Step 5: Calculate $V_x$ at $t = 1.5 s$.
$V_x(1.5) = 2.5 e^{-(1.5 - 0.5) / 2} = 2.5 e^{-1 / 2} = 2.5 e^{-0.5} \approx 2.5 \times 0.6065 = 1.51625 V$. 
Step 6: Round off to two decimal places.
$V_x(1.5) \approx 1.52 V$.