In one RC time constant, a capacitor will charge to approximately what percentage of its full charge?

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Multiple Choice

In one RC time constant, a capacitor will charge to approximately what percentage of its full charge?

Explanation:
In one RC time constant, a capacitor will charge to approximately 63.2% of its full charge. This percentage is derived from the mathematical description of the charging process of a capacitor in an RC (Resistor-Capacitor) circuit. During charging, the voltage across the capacitor does not increase instantaneously; rather, it follows an exponential curve governed by the time constant (τ), which is calculated as the product of the resistance (R) and the capacitance (C). The voltage across the capacitor (Vc) at any time (t) can be expressed with the equation: \[ Vc(t) = Vmax \times (1 - e^{-t/τ}) \] where \( Vmax \) is the maximum voltage (or full charge voltage), and \( e \) is the base of the natural logarithm. After one time constant (t = τ), \( Vc(t) \) equals \( Vmax \times (1 - e^{-1}) \). Since \( e^{-1} \) is approximately 0.3679, this simplifies to about 0.6321 or 63.2% of the maximum charge. The significance of knowing this percentage lies in practical applications

In one RC time constant, a capacitor will charge to approximately 63.2% of its full charge. This percentage is derived from the mathematical description of the charging process of a capacitor in an RC (Resistor-Capacitor) circuit.

During charging, the voltage across the capacitor does not increase instantaneously; rather, it follows an exponential curve governed by the time constant (τ), which is calculated as the product of the resistance (R) and the capacitance (C). The voltage across the capacitor (Vc) at any time (t) can be expressed with the equation:

[ Vc(t) = Vmax \times (1 - e^{-t/τ}) ]

where ( Vmax ) is the maximum voltage (or full charge voltage), and ( e ) is the base of the natural logarithm.

After one time constant (t = τ), ( Vc(t) ) equals ( Vmax \times (1 - e^{-1}) ). Since ( e^{-1} ) is approximately 0.3679, this simplifies to about 0.6321 or 63.2% of the maximum charge.

The significance of knowing this percentage lies in practical applications

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