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An uncharged 10-μf capacitor and a 470-kω resistor are connected in series, and 250 v is applied across the combination.

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Final answer:

To calculate the time taken for a capacitor to reach 110 V, we apply the formula for the voltage across a charging capacitor and solve for time using the resistor and capacitor values given. The result is approximately 2721.3 seconds.

Step-by-step explanation:

The student is asking about the charging of a capacitor in an RC circuit, which is a physics concept related to electricity and circuits. To find out how long it takes the capacitor voltage to reach 110 V, we use the formula for the voltage across a charging capacitor:

V(t) = V0(1 - e-t/RC)

Where:

  • V(t) is the voltage across the capacitor at time t
  • V0 is the initial voltage across the capacitor (250 V in this case)
  • R is the resistance (470 kΩ)
  • C is the capacitance (10 μF)
  • t is the time in seconds
  • e is the base of the natural logarithm (approximately 2.71828)

To find the time t when the voltage is 110 V, we can rearrange the formula to solve for t:

t = -RC ln(1 - ⅓)

The RC time constant (τ) for the circuit is R multiplied by C:

τ = R ⋅ C

τ = 470 kΩ ⋅ 10 μF = 4700 seconds

Now, plug the values into the rearranged formula:

t = -4700 s ⋅ ln(1 - 110 V / 250 V)

t ≈ 4700 s ⋅ ln(1 -0.44)

t ≈ 4700 s ⋅ ln(0.56)

t ≈ 4700 s ⋅ (-0.579)

t ≈ 2721.3 s

Therefore, it takes approximately 2721.3 seconds for the capacitor voltage to reach 110 V.

User Antony Carthy
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