You charge an initially uncharged 82.5mF capacitor through a 42.3Ω resistor by means of a 9.00 V battery having negligible internal resistance. Find the time constant τ of the circuit. What is the charge Q on the capacitor 1.79 time constants after the circuit is closed? What is the charge Q₀ after a long amount of time has passed?
The Correct Answer and Explanation is:
Step 1: Find the time constant, τ
The time constant ( \tau ) for an RC (Resistor-Capacitor) circuit is given by the formula:
[
\tau = R \times C
]
Where:
- ( R ) is the resistance in ohms (( \Omega )),
- ( C ) is the capacitance in farads (F).
Given:
- ( R = 42.3 \, \Omega ),
- ( C = 82.5 \, \text{mF} = 82.5 \times 10^{-3} \, \text{F} ).
Now, substitute these values into the formula:
[
\tau = (42.3 \, \Omega) \times (82.5 \times 10^{-3} \, \text{F}) = 3.49 \, \text{seconds}.
]
Step 2: Calculate the charge on the capacitor after 1.79 time constants
The charge ( Q(t) ) on the capacitor at any time ( t ) during the charging process is given by:
[
Q(t) = Q_{\text{max}} \left( 1 – e^{-t/\tau} \right)
]
Where:
- ( Q_{\text{max}} ) is the maximum charge the capacitor can hold (after a long time),
- ( t ) is the time in seconds,
- ( \tau ) is the time constant.
We need to find the charge after 1.79 time constants, so ( t = 1.79 \tau ).
The maximum charge ( Q_{\text{max}} ) on the capacitor is given by:
[
Q_{\text{max}} = C \times V
]
Where:
- ( C ) is the capacitance,
- ( V ) is the voltage of the battery.
Given:
- ( V = 9.00 \, \text{V} ),
- ( C = 82.5 \times 10^{-3} \, \text{F} ).
The maximum charge ( Q_{\text{max}} ) is:
[
Q_{\text{max}} = (82.5 \times 10^{-3} \, \text{F}) \times (9.00 \, \text{V}) = 0.7425 \, \text{C}.
]
Now, substitute ( t = 1.79 \tau = 1.79 \times 3.49 \, \text{s} = 6.25 \, \text{s} ) into the equation for ( Q(t) ):
[
Q(t) = 0.7425 \, \text{C} \left( 1 – e^{-6.25/3.49} \right)
]
First, calculate the exponent:
[
\frac{6.25}{3.49} \approx 1.79
]
Now, calculate the charge:
[
Q(t) = 0.7425 \, \text{C} \left( 1 – e^{-1.79} \right) = 0.7425 \, \text{C} \times (1 – 0.167) = 0.7425 \, \text{C} \times 0.833 = 0.619 \, \text{C}.
]
Thus, the charge on the capacitor after 1.79 time constants is approximately ( Q(t) = 0.619 \, \text{C} ).
Step 3: Find the charge after a long time (steady state)
After a long amount of time, the capacitor is fully charged, and the current stops. In this steady state, the charge on the capacitor is equal to ( Q_{\text{max}} ).
Thus, the charge on the capacitor after a long time is:
[
Q_{\infty} = Q_{\text{max}} = 0.7425 \, \text{C}.
]
Summary of Answers:
- The time constant ( \tau = 3.49 \, \text{seconds} ),
- The charge on the capacitor after 1.79 time constants is ( Q(t) \approx 0.619 \, \text{C} ),
- The charge on the capacitor after a long time is ( Q_{\infty} = 0.7425 \, \text{C} ).