Thank you for visiting A 605 μF capacitor is discharged through a resistor and its potential difference decreases from an initial value of 96 7 V to 19 9. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Final answer:
The resistance of the resistor is approximately 39.4 kΩ (option b).
Explanation:
Let’s find the resistance of the resistor in this capacitor discharge scenario.
We’ll use the following information:
Capacitance (C) = 605 μF (which we’ll convert to farads: (1 ,[tex]\mu F = 10^{-6}[/tex] , F)
Initial potential difference (V_0) = 96.7 V
Final potential difference (V) = 19.9 V
Time (t) = 4.13 s
The discharge of a capacitor through a resistor is described by the equation:
[tex]$[ V = V_0 \cdot e^{-\frac{t}{RC}} ][/tex]
where:
(V) is the potential difference at time (t)
(R) is the resistance
(C) is the capacitance
(e) is the base of natural logarithm (approximately 2.71828)
We can rearrange the equation to solve for the resistance:
[tex]$[ R = -\frac{t}{C \cdot \ln\left(\frac{V}{V_0}\right)} ][/tex]
Now let’s calculate:
Convert the capacitance to farads: [ C = 605 [tex], \mu F = 605 \times 10^{-6}[/tex] , F = 6.05 \times 10^{-4} , F ]
Calculate the natural logarithm term:[tex][ \ln\left(\frac{V}{V_0}\right) = \ln\left(\frac{19.9}{96.7}\right) ][/tex]
Find the resistance: [tex][ R = -\frac{4.13}{6.05 \times 10^{-4} \cdot \ln\left(\frac{19.9}{96.7}\right)} ][/tex]
Using a calculator, we get: [ R ≈ 39.4 ,[tex]k\Omega[/tex] ]
Correct answer: b) 39.4 kΩ
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Rewritten by : Jeany
Final answer:
In this case, the resistance of the resistor in kilohms is approximately 4.32 kΩ.
So, none of the given options is correct
Explanation:
The provided solution involves the calculation of resistance for a discharging capacitor using the formula [tex]\[ R = \frac{t}{C(\ln(V_0) - \ln(V))} \][/tex]. However, upon following the calculations step by step, it appears there might be an error in the final result.
1. Given values:
- Time (t) = 4.13 seconds
- Capacitance (C) = 605 μF = 605 x 10â»â¶ F
- Initial voltage (Vâ‚€) = 96.7 volts
- Final voltage (V) = 19.9 volts
2. Substitute the values into the formula:
[tex]\[ R = \frac{4.13}{605 x 10^{-6} (\ln(96.7) - \ln(19.9))} \][/tex]
3. Perform the calculations step by step:
[tex]\[ R = \frac{4.13}{0.000605 (\ln(96.7) - \ln(19.9))} \][/tex]
[tex]\[ R = \frac{4.13}{0.000605 (\ln(96.7/19.9))} \][/tex]
[tex]\[ R = \frac{4.13}{0.000605 (\ln(4.854))} \][/tex]
[tex]\[ R = \frac{4.13}{0.000605 \times 1.5777} \][/tex]
[tex]\[ R = \frac{4.13}{0.0009556985} \][/tex]
[tex]\[ R = 4316.25 \, \Omega \][/tex]
4. Convert the resistance into kilohms:
[tex]\[ R = \frac{4316.25}{1000} \][/tex]
[tex]\[ R = 4.31625 \, k\Omega \][/tex]
[tex]\[ R = 4.32\, k\Omega \] ( Rounded up)[/tex]
Therefore, the calculated resistance for the discharging capacitor is approximately 4.32 kΩ
So, none of the given options is correct
