How long will it take for a 50 µF capacitor to fully charge if it is in a series circuit with a 100 KΩ resistor?
The Correct Answer and Explanation is :
The correct answer is: 25 seconds
To determine how long it will take for a 50 µF capacitor to fully charge in a series circuit with a 100 kΩ resistor, we can use the time constant formula of an RC (resistor-capacitor) circuit. The time constant (( \tau )) is given by:
[
\tau = R \times C
]
Where:
- ( \tau ) is the time constant in seconds,
- ( R ) is the resistance in ohms,
- ( C ) is the capacitance in farads.
Given:
- ( R = 100,000 \, \Omega ) (100 kΩ)
- ( C = 50 \, \mu F = 50 \times 10^{-6} \, F )
Calculation of the Time Constant:
- Convert capacitance to farads:
[
C = 50 \times 10^{-6} \, F
] - Plug in the values:
[
\tau = 100,000 \, \Omega \times 50 \times 10^{-6} \, F = 5 \, seconds
]
Charging Time:
The capacitor charges approximately to 63.2% of the supply voltage after one time constant (( \tau )). The charging curve of the capacitor can be described by the equation:
[
V(t) = V_{max} \left(1 – e^{-t/\tau}\right)
]
Where:
- ( V(t) ) is the voltage across the capacitor at time ( t ),
- ( V_{max} ) is the maximum voltage (supply voltage),
- ( e ) is the base of the natural logarithm (approximately 2.718).
To consider the time it takes to “fully charge” (approaching ( V_{max} )), it’s often considered that after about ( 5\tau ) (or 5 time constants), the capacitor will be over 99% charged.
Final Calculation:
For our case:
[
5 \tau = 5 \times 5 \, seconds = 25 \, seconds
]
Conclusion:
Thus, it will take approximately 25 seconds for the 50 µF capacitor to fully charge in a series circuit with a 100 kΩ resistor. This calculation assumes ideal conditions without any significant losses, allowing the capacitor to reach a near-complete charge relative to the applied voltage.