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A [tex]563 \times 10^{-6} \, \text{F}[/tex] capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of [tex]99.3 \, \text{V}[/tex] to [tex]25.1 \, \text{V}[/tex] in [tex]4.21 \, \text{s}[/tex].

Find the resistance [tex]R[/tex] of the resistor.

Answer :

Resistance (R) of the resistor ≈ 2443.89 ohms.

Using the formula R = -t / (C * ln(V(t) / V₀)).

values: V₀ = 99.3 V, V(t) = 25.1 V, t = 4.21 s, C = 563×10^(-6) F.

To find the resistance (R) of the resistor, we can use the formula for the discharge of a capacitor in an RC circuit:

V(t) = V₀ * e^(-t/RC),

where V(t) is the voltage at time t, V₀ is the initial voltage, R is the resistance, C is the capacitance, and e is the base of the natural logarithm.

In this case, we are given the following information:

- Initial voltage, V₀ = 99.3 V

- Final voltage, V(t) = 25.1 V

- Time, t = 4.21 s

- Capacitance, C = 563×10^(-6) F

We can rearrange the formula to solve for the resistance R:

R = -t / (C * ln(V(t) / V₀)).

Plugging in the values:

R = -4.21 s / (563×10^(-6) F * ln(25.1 V / 99.3 V)) ≈ 2443.89 Ω.

Therefore, the resistance of the resistor is approximately 2443.89 ohms.

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