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Suppose you wanted to to discharge a 150 μF capacitor through a 350 Ω resistor down to 1.00% of its original voltage. Randomized Variables R = 350 Ω C = 150 μF. How much time would be required in s?

User Slashterix
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1 Answer

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Final answer:

It would take approximately 5.258 seconds to discharge a 150 μF capacitor through a 350 Ω resistor down to 1.00% of its original voltage.

Step-by-step explanation:

The time required to discharge a capacitor through a resistor can be calculated using the formula for the RC time constant and the exponential decay function. Specifically, the voltage across a discharging capacitor as a function of time, V(t), is given by:

V(t) = V_0 × e^{-t/RC}

where:

  • V_0 is the initial voltage,
  • e is the base of the natural logarithm,
  • t is the time in seconds,
  • R is the resistance in ohms (Ω), and
  • C is the capacitance in farads (F).

To find the time when the voltage is at 1.00% of its original value, we can rearrange the equation to solve for t:

t = -RC × ln(0.01)

For the given values, R = 350 Ω and C = 150 μF, we have:

t = -350 × 150×10^-6 × ln(0.01) ≈ 5.258 seconds

Therefore, it would take approximately 5.258 seconds to discharge the 150 μF capacitor through a 350 Ω resistor down to 1.00% of its original voltage.

User Walterfaye
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