Solutions for the Capacitor Experiment #7A

1. Calculate the time constant ($\tau_c$):

   The time constant is calculated using: $\tau_c = R \cdot C$ Given: $R = 20,000 \, \Omega \quad \text{and} \quad C = 470 \, \mu F = 470 \times 10^{-6} \, F$ $\tau_c = 20,000 \cdot 470 \times 10^{-6} = 9.4 \, \text{seconds}$

2. Calculate the voltage across the capacitor at $t = \tau_c$:

   The voltage at time $t$ for a charging capacitor is given by: $V_c(t) = V \cdot \left( 1 - e^{-t / \tau_c} \right)$ At $t = \tau_c$: $V_c = 12 \cdot \left( 1 - e^{-1} \right) = 12 \cdot \left( 1 - 0.3679 \right) = 12 \cdot 0.6321 = 7.5852 \, \text{volts}$

3. Calculate the charging time ($5\tau$):

   The capacitor is considered fully charged after $5\tau$: $t_{\text{charge}} = 5 \cdot \tau_c = 5 \cdot 9.4 = 47 \, \text{seconds}$

4. Initial voltage across the capacitor ($V_{c,\text{init}}$):

   Initially, before charging starts ($t = 0$): $V_{c,\text{init}} = 0 \, \text{volts}$

5. Final voltage across the capacitor ($V_{c,\text{final}}$):

   Once fully charged, the capacitor voltage equals the source voltage: $V_{c,\text{final}} = 12 \, \text{volts}$

6. Initial current in the circuit ($I_c$):

   The initial current is determined by Ohm’s law: $I_c = \frac{V}{R}$ $I_c = \frac{12}{20,000} = 0.0006 \, \text{amperes} = 0.6 \, \text{mA}$

Final Answers:

1. $\tau_c = 9.4 \, \text{seconds}$
2. $V_c = 7.59 \, \text{volts}$
3. $t_{\text{charge}} = 47 \, \text{seconds}$
4. $V_{c,\text{init}} = 0 \, \text{volts}$
5. $V_{c,\text{final}} = 12 \, \text{volts}$
6. $I_c = 0.6 \, \text{mA}$

Would you like further explanations or clarifications?

Related Questions:

1. How does the resistance value affect the time constant $\tau_c$?
2. What happens to the charging time if the capacitance is doubled?
3. How can you experimentally verify the theoretical values of $\tau_c$?
4. Why does the capacitor voltage never exceed the source voltage?
5. What is the physical significance of $5\tau$ in the charging process?

Tip:

Always use a multimeter to measure both resistance and capacitance accurately before calculations for more precise