A circuit contains a voltage source Vs = 1 V, a resistor R1 = 10 kΩ, a resistor R2, and a capacitor C = 4.7 μF. The source frequency is 10 Hz. Find the output voltage across C when (a) R1 is open, and

Step 1: Calculate the capacitive reactance.
The source frequency is $f = 10\,Hz$ and the capacitance is $C = 4.7\,\mu F$.
First, calculate the angular frequency $\omega$:

$$\omega = 2\pi f = 2\pi (10\,Hz) = 20\pi\,rad/s$$

Next, calculate the capacitive reactance $X_C$:

$$X_C = \frac{1}{\omega C} = \frac{1}{20\pi \times 4.7 \times 10^{-6}\,F}$$ $$X_C = \frac{1}{295.3097 \times 10^{-6}}\,\Omega \approx 3386.3\,\Omega$$

The impedance of the capacitor is $Z_C = -jX_C = -j3386.3\,\Omega$.

Step 2: Determine the output voltage for (a) R1 open.
If $R_1$ is open, it creates an open circuit in series with the voltage source. No current can flow through the circuit.
Since no current flows through $R_2$ or $C$, there will be no voltage drop across them.
Therefore, the output voltage $V_{out}$ is $\boxed{\text{0 V}}$.

Step 3: Determine the output voltage for (b) R2 open.
If $R_2$ is open, the resistor $R_2$ is removed from the circuit. The circuit becomes a series combination of $R_1$ and $C$.
The total impedance of this series circuit is $Z_{total} = R_1 + Z_C = 10\,k\Omega - j3386.3\,\Omega$.
The output voltage $V_{out}$ is measured across the capacitor $C$. Using the voltage divider rule:

$$V_{out} = V_s \frac{Z_C}{R_1 + Z_C}$$ $$V_{out} = 1\,V \frac{-j3386.3\,\Omega}{10000\,\Omega - j3386.3\,\Omega}$$

The magnitude of the output voltage is:

$$|V_{out}| = 1\,V \frac{|-j3386.3\,\Omega|}{|10000\,\Omega - j3386.3\,\Omega|}$$ $$|V_{out}| = \frac{3386.3}{\sqrt{(10000)^2 + (-3386.3)^2}}$$ $$$$