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A 2.8 F capacitor and a 3,440
Ω
resistor are connected to a battery of voltage 5 V as shown in the circuit. After
closing the switch, how long will it take for the capacitor voltage to be 45 % of
the battery voltage?
Express your answer in seconds (s)

Respuesta :

onyebuchinnaji

The time taken for the capacitor voltage to be 45 % of the battery voltage is 5,750.3 s.

Discharge of the capacitor voltage

The rate at which the capacitor voltage discharges is given by the following formula;

[tex]V = V_0(1 - e^{-t/RC})[/tex]

Where;

  • V₀ is the initial voltage
  • V is the final voltage
  • t is the time of discharge
  • R is resistance
  • C is capacitance

The final voltage = 0.45 x 5 V = 2.25 V

[tex]2.25 = 5(1 - e^{-t/RC})\\\\(1 - e^{-t/RC}) = \frac{2.25}{5} \\\\(1 - e^{-t/RC}) = 0.45\\\\e^{-t/RC} = 1- 0.45\\\\e^{-t/RC} = 0.55\\\\\frac{-t}{RC} = ln(0.55)\\\\\frac{-t}{RC} = -0.597\\\\t = 0.597 (RC)\\\\t = 0.597(3440 \times 2.8)\\\\t = 5,750.3 \ s[/tex]

Thus, the time taken for the capacitor voltage to be 45 % of the battery voltage is 5,750.3 s.

Learn more about capacitor voltage here: https://brainly.com/question/14883923

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