This question asks about the voltage across a capacitor after it has been charging from a 10 V DC source for a duration equal to one time constant ($\tau$).

Understanding Capacitor Charging

When a capacitor charges through a resistor (forming an RC circuit) from a DC voltage source, the voltage across the capacitor, denoted as $V(t)$, increases over time. The rate of charging is determined by the time constant, $\tau$, which is the product of the resistance (R) and the capacitance (C) in the circuit ($\tau = RC$).

The formula for the voltage across a charging capacitor is:

$$ V(t) = V_s \left(1 - e^{-t/\tau}\right) $$

Where:

• $V(t)$ is the voltage across the capacitor at time $t$.
• $V_s$ is the source voltage.
• $t$ is the time elapsed since charging began.
• $\tau$ is the time constant of the circuit.
• $e$ is the base of the natural logarithm (approximately 2.71828).

Calculating Voltage at One Time Constant

We need to find the voltage across the capacitor when the charging time $t$ is equal to one time constant, so $t = \tau$. Substituting this into the formula:

$$ V(\tau) = V_s \left(1 - e^{-\tau/\tau}\right) $$

Simplifying the exponent:

$$ V(\tau) = V_s \left(1 - e^{-1}\right) $$

Now, we use the given values:

• Source Voltage, $V_s = 10$ V.
• Time, $t = \tau$.

Plugging $V_s = 10$ V into the equation:

$$ V(\tau) = 10 \text{ V} \left(1 - e^{-1}\right) $$

We know that $e^{-1}$ is approximately 0.36788. Calculating the value:

$$ V(\tau) \approx 10 \text{ V} \left(1 - 0.36788\right) $$

$$ V(\tau) \approx 10 \text{ V} \left(0.63212\right) $$

$$ V(\tau) \approx 6.3212 \text{ V} $$

Conclusion

After one time constant, the voltage across the capacitor reaches approximately 63.2% of the source voltage. In this case, with a 10 V source, the voltage is approximately 6.32 V. Comparing this to the given options, the closest value is 6.3 V.